{"id":626471,"date":"2023-09-12T02:36:00","date_gmt":"2023-09-12T02:36:00","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=626471"},"modified":"2023-09-12T02:36:20","modified_gmt":"2023-09-12T02:36:20","slug":"kc-sinha-exercise-19-3-mathematics-solution-class-12-chapter-19-indefinite-integrals","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-19-3-mathematics-solution-class-12-chapter-19-indefinite-integrals\/","title":{"rendered":"KC Sinha: Exercise 19.3- Mathematics Solution Class 12 Chapter 19 Indefinite Integrals"},"content":{"rendered":"\n

<\/span><\/p>\n\n\n\n

Exercise 19.3<\/h3>\n\n\n\n

Question 1<\/strong>
$\\displaystyle\\int(x+2) \\sqrt{3 x+5} d x$
Sol :
Let $z=3 x+5$ then $\\dfrac{d z}{d x}=3 $<\/p>\n\n\n\n

\u2234$d x=\\dfrac{d z}{3}$<\/p>\n\n\n\n

Again z=3 x+5 \u2234 $x=\\dfrac{2-5}{3}$<\/p>\n\n\n\n

Now, $\\displaystyle\\int(x+2) \\cdot \\sqrt{3 x+5} d x$<\/p>\n\n\n\n

$=\\displaystyle\\int\\left(\\dfrac{z-5}{3}+2\\right) \\cdot \\sqrt{z}  \\dfrac{dz}{3}$<\/p>\n\n\n\n

$=\\frac{1}{3} \\displaystyle\\int\\left(\\frac{z-5+6}{3}\\right) \\sqrt{z} d z$<\/p>\n\n\n\n

$=\\frac{1}{9} \\displaystyle\\int(z+1) \\sqrt{z} d z$<\/p>\n\n\n\n

$=\\dfrac{1}{9} \\displaystyle\\int z^{\\tfrac{3}{2}} dz+\\dfrac{1}{9} \\displaystyle\\int z^{\\tfrac{1}{2}} d z$<\/p>\n\n\n\n

$=\\dfrac{1}{9} \\cdot \\dfrac{2}{5} \\cdot z^{\\dfrac{5}{2}}+\\dfrac{1}{9} \\cdot \\dfrac{2}{3} \\cdot z^{\\tfrac{3}{2}}+c$<\/p>\n\n\n\n

=$z^{\\tfrac{3}{2}}\\left[\\dfrac{2}{45}z+\\dfrac{2}{27}\\right]$+c<\/p>\n\n\n\n

=$z^{\\tfrac{3}{2}}\\left[\\dfrac{6z+10}{135}\\right]$+c<\/p>\n\n\n\n

=$\\dfrac{1}{135}.(3x+5)^{\\tfrac{3}{5}}.\\left[6(3x+5)+10\\right]$+c<\/p>\n\n\n\n

=$\\dfrac{1}{135}.(3x+5)^{\\tfrac{3}{5}}.\\left[18x+40\\right]$+c<\/p>\n\n\n\n

=$\\dfrac{2}{135}.(9x+20).(3x+5)^{\\tfrac{3}{2}}$+c<\/p>\n\n\n\n

Question 2<\/strong>
$\\displaystyle\\int (x-1).\\sqrt{x+1}dx$
Sol :
Let z=x+1 then $\\dfrac{dz}{dx}=1$<\/p>\n\n\n\n

\u2234dx=dz<\/p>\n\n\n\n

Again , z=x+1 \u2234x=z-1<\/p>\n\n\n\n

Now, $\\displaystyle\\int (x-1)\\sqrt{x+1}dx$<\/p>\n\n\n\n

=$\\displaystyle\\int (z-1-1).\\sqrt{z}dz$<\/p>\n\n\n\n

=$\\displaystyle\\int (z-2).\\sqrt{z}dz$<\/p>\n\n\n\n

=$\\displaystyle\\int z^{\\tfrac{3}{5}}dz-2\\displaystyle\\int z^{\\tfrac{1}{2}}dz$<\/p>\n\n\n\n

=$\\dfrac{2}{5}.z^{\\tfrac{5}{2}}-2.\\dfrac{2}{3}.z^{\\tfrac{3}{2}}$+c<\/p>\n\n\n\n

=$\\dfrac{2}{5}.z^{\\tfrac{5}{2}}-\\dfrac{4}{3}.z^{\\tfrac{3}{2}}$+c<\/p>\n\n\n\n

=$z^{\\tfrac{3}{2}}\\left[\\dfrac{2}{5}z-\\dfrac{4}{3}\\right]$+c<\/p>\n\n\n\n

=$z^{\\tfrac{3}{2}}\\left[\\dfrac{6z-20}{15}\\right]$+c<\/p>\n\n\n\n

=$z^{\\tfrac{3}{2}}\\left[\\dfrac{6(x+1)-20}{15}\\right]$+c<\/p>\n\n\n\n

=$(x+1)^{\\tfrac{3}{2}}.\\left(\\dfrac{6x+6-20}{15}\\right)$+c<\/p>\n\n\n\n

=$(x+1)^{\\tfrac{3}{2}}.2\\left(\\dfrac{3x-7}{15}\\right)$+c<\/p>\n\n\n\n

=$\\dfrac{2}{15}.(x+1)^{\\tfrac{3}{2}}.(3x-7)$+c<\/p>\n\n\n\n

Question 3<\/strong>
$\\int(x+2) \\cdot \\sqrt{2 x+1} d x$
Sol:
Let $z=2 x+1$ then $\\frac{d z}{d x}=2 \\therefore d x=\\frac{d z}{2}$<\/p>\n\n\n\n

again, z=2 x+1 $\\therefore x=\\frac{z-1}{2}$<\/p>\n\n\n\n

Now, $\\left.\\int\\left(\\frac{z-1}{2}\\right)+2\\right) \\cdot \\sqrt{z} \\frac{d z}{2}=\\int\\left(\\frac{z-1+4}{2}\\right) \\sqrt{z} \\frac{d z}{2}$<\/p>\n\n\n\n

$=\\frac{1}{4} \\int(z+3) \\cdot \\sqrt{z} d z$<\/p>\n\n\n\n

$=\\frac{1}{4} \\int z^{\\frac{3}{2}} d z+\\frac{3}{4} \\int z^{\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=\\frac{1}{4} \\cdot \\frac{2}{5} \\cdot z^{\\frac{5}{2}}+\\frac{3}{4} \\cdot \\left(\\dfrac{z}{\\frac{3}{2}}\\right)^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

$=\\dfrac{1}{10}z^{\\frac{5}{2}}+\\dfrac{1}{2}z^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

=$z^{\\frac{3}{2}}\\left[\\dfrac{1}{10}z+\\dfrac{1}{2}\\right]+c$<\/p>\n\n\n\n

=$(2x+1)^{\\frac{3}{2}}\\left[\\dfrac{1}{10}(2x+1)+\\dfrac{1}{2}\\right]+c$<\/p>\n\n\n\n

=$(2x+1)^{\\frac{3}{2}}\\left[\\dfrac{2x+1+5}{10}\\right]+c$<\/p>\n\n\n\n

=$(2x+1)^{\\frac{3}{2}}\\left[\\dfrac{2x+6}{10}\\right]+c$<\/p>\n\n\n\n

=$\\dfrac{2}{10}(x+3)(2x+1)^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

Question 4<\/strong>
$\\int x \\cdot \\sqrt{2 x+1} d x$
Sol :
Let z=2x+1 , then $\\frac{d z}{d x}=2$ $\\therefore d x=\\frac{d z}{2}$<\/p>\n\n\n\n

Again Z=2x+1 $\\therefore x=\\frac{z-1}{2}$<\/p>\n\n\n\n

Now , $\\int x \\cdot \\sqrt{2 x+1} d x$<\/p>\n\n\n\n

$=\\int \\frac{z-1}{2} \\cdot \\sqrt{z} \\frac{d z}{2}$<\/p>\n\n\n\n

$=\\frac{1}{4} \\int z^{\\frac{3}{2}}-z^{\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=\\frac{1}{4} \\int z^{\\frac{3}{2}} d z-\\frac{1}{4} \\int z^{\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=\\frac{1}{4} \\int z^{\\frac{3}{2}}-z^{\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=\\frac{1}{4} \\int z^{\\frac{3}{2}} d z-\\frac{1}{4} \\int z^{\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=\\frac{1}{4} \\cdot \\frac{2}{5} \\cdot z^{\\frac{5}{2}}-\\frac{1}{4} \\cdot \\frac{2}{3} \\cdot z^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

$=\\frac{1}{10} z^{\\frac{5}{2}}-\\frac{1}{6} z^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

$=z^{\\frac{3}{2}}\\left[\\frac{1}{10} z-\\frac{1}{6}\\right]+c$<\/p>\n\n\n\n

$=z^{\\frac{3}{2}}\\left[\\frac{2 x+1}{10}-\\frac{1}{6}\\right]+c$<\/p>\n\n\n\n

$=(2 x+1)^{\\frac{3}{2}} \\cdot\\left(\\frac{6 x-2}{30}\\right)+c$<\/p>\n\n\n\n

$=\\frac{2(3 x-1)}{30} \\cdot(2 x+1)^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

$=\\frac{1}{15}(3 x-1) \\cdot(2 x+1)^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

Question 5<\/strong>
$\\int x \\cdot(3 x-5)^{4} d x$
Sol :
Let z=3x-5 then $\\frac{d z}{d x}=3 \\therefore d x=\\frac{d z}{3}$<\/p>\n\n\n\n

Again z=3x-5 $\\therefore x=\\dfrac{z+5}{3}$<\/p>\n\n\n\n

Now $\\int x \\cdot(3 x-5)^{4} d x$ $=\\int \\frac{z+5}{3} \\cdot z^{4} \\frac{d z}{3}$<\/p>\n\n\n\n

$=\\frac{1}{9} \\int z^{5} d z+\\frac{5}{9} \\int z^{4} d z$<\/p>\n\n\n\n

$=\\frac{1}{9} \\cdot \\frac{z^{6}}{6}+\\frac{5}{9} \\cdot \\frac{z^{5}}{5}+c$<\/p>\n\n\n\n

$=z^{5}\\left[\\frac{2}{5}+\\frac{1}{5}\\right]+c$<\/p>\n\n\n\n

$=z^{5}\\left[\\frac{z+6}{54}\\right]+c$<\/p>\n\n\n\n

$=\\frac{1}{54} \\cdot(3 x-5)^{5} \\cdot(3 x+1)+c$<\/p>\n\n\n\n

Question 11<\/strong>
$\\int x^{2} \\cdot \\sqrt{x+2} d x$
Sol :
Let z=x+2 \\therefore \\frac{d z}{d x}=1 \\quad \\therefore \\quad d z=d x
Again z=x+2 , x=z-2<\/p>\n\n\n\n

Now $\\int x^{2} \\cdot \\sqrt{x+2} d x=\\int(z-2)^{2} \\cdot \\sqrt{z} d z$<\/p>\n\n\n\n

$=\\int\\left(z^{2}+4-4 z\\right) \\sqrt{z} d z$<\/p>\n\n\n\n

$=\\int z^{\\frac{1}{2}} d z+4 \\int z^{\\frac{1}{2}} d z-4 \\int z^{\\frac{3}{2}} d z$<\/p>\n\n\n\n

$=\\frac{2}{7} \\cdot 2^{\\frac{7}{2}}+4 \\cdot \\frac{2}{3} \\cdot z^{\\frac{3}{2}}-4 \\cdot \\frac{2}{5} \\cdot z^{\\frac{5}{2}}+c$<\/p>\n\n\n\n

$=2 z^{\\frac{3}{2}}\\left[\\frac{1}{7} z^{2}+\\frac{4}{3}-\\frac{4}{5} z\\right]+c$<\/p>\n\n\n\n

=2(x+2)^{\\frac{3}{2}}\\left(\\frac{(x+2)^{2}}{7}-\\frac{4}{5}(x+2)+\\frac{4}{3}\\right]+c<\/p>\n\n\n\n

Question 12<\/strong>
$\\int\\left(2 x^{2}+3\\right) \\sqrt{x+2} d x$
Sol :
Let z=x+2<\/p>\n\n\n\n

$\\therefore \\frac{d z}{d x}=1 \\quad \\therefore \\quad d z=d x$<\/p>\n\n\n\n

again z=x+2 \u2234x=z-2<\/p>\n\n\n\n

Now , $\\int\\left(2 x^{2}+3\\right) \\cdot \\sqrt{x+2} d x$<\/p>\n\n\n\n

$=\\int\\left\\{2(z-2)^{2}+3\\right\\} \\sqrt{z} d 2$<\/p>\n\n\n\n

$=\\int\\left\\{2\\left(z^{2}+4-4z\\right)+3\\right\\} \\sqrt{2} d z$<\/p>\n\n\n\n

$=\\int\\left(2 z^{2}+8-8 z+3\\right) \\sqrt{z} d z$<\/p>\n\n\n\n

$=\\int\\left(2z^{2}+11-8z\\right) \\sqrt{z} d 2$<\/p>\n\n\n\n

$=2 \\int2^{\\frac{5}{2}} d z+11 \\int z^{\\frac{1}{2}} d z-8\\int z^{\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=2 \\cdot \\frac{2}{7} \\cdot z^{\\frac{7}{2}}+11 \\cdot \\frac{2}{3} \\cdot z^{\\frac{3}{2}}-8 \\cdot \\frac{2}{5} \\cdot z^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

$=2z^{\\frac{3}{2}}\\left[\\frac{2}{7} z^{2}+\\frac{11}{3}-\\frac{8}{5} z\\right]+c$<\/p>\n\n\n\n

$=2(x+2)^{\\frac{3}{2}}\\left[\\frac{2}{7}(x+8)^{2}-\\frac{8}{5}(x+2)+\\frac{11}{3}\\right]+c$<\/p>\n\n\n\n

Question 13<\/strong>
$\\int \\frac{x}{\\sqrt{x+5}} d x$
Sol :
Let x+5=z then $\\frac{d z}{d x}=1$ \u2234dx=dz<\/p>\n\n\n\n

Again , x+5=z  \u2234x=z-5<\/p>\n\n\n\n

Now , $\\int \\frac{x}{\\sqrt{x+5}} d x$<\/p>\n\n\n\n

$=\\int \\frac{z-5}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\int \\frac{z}{\\sqrt{z}} d z-5 \\int \\frac{1}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\int \\sqrt{z} d z-5 \\int z^{-\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=\\frac{2}{3} \\cdot z^{\\frac{3}{2}}-5 \\times 2 \\cdot z^{-\\frac{1}{2}}+c$<\/p>\n\n\n\n

$=z^{\\frac{1}{2}}\\left[\\frac{2}{3} z-10\\right]+c$<\/p>\n\n\n\n

$=\\sqrt{x+5} \\cdot\\left(\\frac{2}{3}(x+5)-10\\right)+c$<\/p>\n\n\n\n

$=\\sqrt{x+5}.\\left(\\frac{2 x+10-30}{3}\\right)+c$<\/p>\n\n\n\n

$=\\sqrt{x+5} \\cdot \\frac{(2 x-20)}{3}+c$<\/p>\n\n\n\n

$=\\frac{2}{3}(x-10) \\cdot \\sqrt{x+5}+c$<\/p>\n\n\n\n

Question 14<\/strong>
$\\int \\frac{x}{\\sqrt{1-2 x}} d x$
Sol :
Let z=1-2x then $\\frac{d z}{d x}=-2 \\therefore d x=-\\frac{d z}{2}$<\/p>\n\n\n\n

Again , z=1-2x $\\therefore x=\\frac{1-z}{2}$<\/p>\n\n\n\n

Now , $\\int \\frac{x}{\\sqrt{1-2 x}} \\cdot d x$ $=\\int \\frac{1-2}{2 \\cdot \\sqrt{2}} \\cdot\\left(-\\frac{d z}{2}\\right)$<\/p>\n\n\n\n

$=\\frac{-1}{4} \\int \\frac{(1-z)}{\\sqrt{z}} d 2$<\/p>\n\n\n\n

$=\\frac{-1}{4} \\int \\frac{1}{\\sqrt{z}} d z+\\frac{1}{4}\\int\\frac{z}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=-\\frac{1}{4} \\int z^{-\\frac{1}{2}} d z+\\frac{1}{4} \\int \\sqrt{z} d z$<\/p>\n\n\n\n

$=-\\frac{1}{4} \\cdot 2 \\cdot z^{\\frac{1}{2}}+\\frac{1}{4} \\cdot \\frac{2}{3} \\cdot z^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

$=-\\frac{1}{2} z^{\\frac{1}{2}}+\\frac{1}{6} z^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

$=2^{\\frac{1}{2}}\\left[-\\frac{1}{2}+\\frac{1}{6} z\\right]+c$<\/p>\n\n\n\n

$=z^{\\frac{1}{2}}\\left(\\frac{-3+2}{6}\\right)$<\/p>\n\n\n\n

$=\\sqrt{1-2 x} \\cdot\\left(\\frac{-3+1-2 x}{6}\\right)+c$<\/p>\n\n\n\n

$=\\sqrt{1-2 x}\\left(\\frac{-2-2 x}{6}\\right)+c$<\/p>\n\n\n\n

$=\\frac{-2}{6}(1+x) \\sqrt{1-2 x}+c$<\/p>\n\n\n\n

$=-\\frac{1}{3}(1+x) \\sqrt{1-2 x}+c$<\/p>\n\n\n\n

Question 15<\/strong>
$\\int x \\cdot \\sqrt{x+2} d x$
Sol :
Let z=x+2 then $\\frac{d z}{d x}=1$ \u2234dz=dx<\/p>\n\n\n\n

Again z=x+2  \u2234x=z-2<\/p>\n\n\n\n

Now , $\\int x \\cdot \\sqrt{x+2} d x$<\/p>\n\n\n\n

$=\\int(z-2) \\cdot \\sqrt{z} d z$<\/p>\n\n\n\n

$=\\int z^{\\frac{3}{2}} d z-2 \\int z^{\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=\\frac{2}{5} \\cdot z^{\\frac{5}{2}}-2 \\cdot \\frac{2}{3} \\cdot z^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

$=2 z^{\\frac{3}{2}}\\left[\\frac{z}{5}-\\frac{2}{3}\\right]+c$<\/p>\n\n\n\n

$=2z^{\\frac{3}{2}}\\left[\\frac{3z-10}{15}\\right]+c$<\/p>\n\n\n\n

$=2 \\cdot(x+2)^{\\frac{3}{2}}\\left[\\frac{3(x+2)-10}{15}\\right]+c$<\/p>\n\n\n\n

$=\\frac{2}{15}(x+2)^{\\frac{3}{2}}\\lceil 3 x-4]+c$<\/p>\n\n\n\n

Question 16<\/strong>
$\\int \\frac{2 x+1}{\\sqrt{3 x+2}} d x$
Sol :
Let z=3x+2 then $\\frac{d z}{d x}=3$ $\\therefore d x=\\frac{d z}{3}$<\/p>\n\n\n\n

Again , z=3x+2  $\\therefore \\quad x=\\frac{z-2}{3}$<\/p>\n\n\n\n

Now , $\\int \\frac{2 x+1}{\\sqrt{3 x+2}} d x$<\/p>\n\n\n\n

$=\\int \\frac{2\\left(\\frac{z-2}{3}\\right)+1}{\\sqrt{z}} \\cdot \\frac{d z}{3}$<\/p>\n\n\n\n

$=\\displaystyle\\int \\dfrac{2(z-2)+3}{3 \\sqrt{z}} \\dfrac{dz}{3}$<\/p>\n\n\n\n

$=\\frac{1}{9} \\int \\frac{(2z-4+3)}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\frac{1}{9} \\int \\frac{(2z-1)}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\frac{1}{9} \\cdot 2 \\int \\frac{z}{\\sqrt{z}} d z-\\frac{1}{9} \\int \\frac{1}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\frac{2}{9} \\cdot \\frac{2}{3} \\cdot z^{\\frac{3}{2}}-\\frac{1}{9} \\cdot 2 \\cdot z^{\\frac{1}{2}}+c$<\/p>\n\n\n\n

$=2z^{\\frac{1}{2}}\\left[\\frac{2}{27} \\cdot z-\\frac{1}{9}\\right]+c$<\/p>\n\n\n\n

$=2 z^{\\frac{1}{2}}\\left[\\frac{2 z-3}{27}\\right]+c$<\/p>\n\n\n\n

$=\\frac{2}{27}(3 x+2)^{\\frac{1}{2}} \\cdot[2(3 x+8)-3]+c$<\/p>\n\n\n\n

$=\\frac{2}{27} \\cdot(3 x+2)^{\\frac{1}{2}}[6 x+4-3]+c$<\/p>\n\n\n\n

$=\\frac{2}{27} \\cdot(6 x+1) \\cdot \\sqrt{3 x+2}+c$<\/p>\n\n\n\n

Question 17<\/strong>
$\\int \\frac{3 x+5}{\\sqrt{7 x+9}} d x$
Sol :
Let z=7x+9 then $\\frac{d z}{d x}=7$ $\\therefore d x=\\frac{d z}{7}$<\/p>\n\n\n\n

Again , z=7x+9 $\\therefore x=\\frac{z-9}{7}$<\/p>\n\n\n\n

Now , $\\int \\frac{3 x+5}{\\sqrt{7 x+9}} d x$<\/p>\n\n\n\n

$=\\int \\frac{3\\frac{(z-9)}{7}}{\\sqrt{z}}+5 \\cdot \\frac{d z}{7}$<\/p>\n\n\n\n

$=\\frac{1}{49} \\int \\frac{3z+8}{\\sqrt{2}} d z$<\/p>\n\n\n\n

$=\\frac{1}{49} \\cdot 3 \\int \\frac{z}{\\sqrt{z}} d z+\\frac{8}{49} \\int \\frac{1}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\frac{3}{49} \\int \\sqrt{z} d 2+\\frac{8}{49} \\int z^{-\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=\\frac{3}{49} \\cdot \\frac{2}{3} \\cdot z^{\\frac{3}{2}}+\\frac{8}{49} \\cdot 2 \\cdot z^{\\frac{1}{2}}+c$<\/p>\n\n\n\n

$=\\frac{2z^{\\frac{1}{2}}}{49}[z+8]+c$<\/p>\n\n\n\n

$=\\frac{2}{49} \\cdot \\sqrt{7 x+9} \\cdot(7 x+9+8)+c$<\/p>\n\n\n\n

$=\\frac{2}{49} \\cdot \\sqrt{7 x+9} \\cdot(7 x+17)+c$<\/p>\n\n\n\n

Question 18<\/strong>
$\\int \\frac{x^{2}}{\\sqrt{x-1}} d x$
Sol :
Let z=x-1 then $\\frac{d 2}{d x}=1 \\therefore d x=d 2$<\/p>\n\n\n\n

Again , z=x-1 $\\therefore x=z+1$<\/p>\n\n\n\n

Now , $\\int \\frac{x^{2}}{\\sqrt{x-1}} d x$<\/p>\n\n\n\n

$=\\int \\frac{(z+1)^{2}}{\\sqrt{Z}} d z$<\/p>\n\n\n\n

$=\\int \\frac{z^{2}+1+2 z}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\int \\frac{z^{2}}{\\sqrt{z}} d z+\\int \\frac{1}{\\sqrt{z}} d z+2 \\int \\frac{z}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$\\int z^{\\frac{3}{2}} d z+\\int z^{-\\frac{1}{2}} d z+2 \\cdot \\int z^{\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=\\frac{2}{5} \\cdot z^{\\frac{5}{2}}+2 \\cdot z^{\\frac{1}{2}}+2 \\cdot \\frac{2}{3} \\cdot z^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

$=2z^{\\frac{1}{2}}\\left[\\frac{z^{2}}{5}+1+\\frac{2}{3} z\\right]+c$<\/p>\n\n\n\n

$=2 \\sqrt{z}\\left[\\frac{3z^{2}+15+10z}{15}\\right]+c$<\/p>\n\n\n\n

$=\\frac{2}{15} \\sqrt{x-1}\\left[3(x-1)^{2}+15+10(x-1)\\right]+c$<\/p>\n\n\n\n

$=\\frac{2}{15} \\cdot \\sqrt{x-1}\\left[3\\left(x^{2}+1-2 x\\right)+15+10 x-10\\right]+c$<\/p>\n\n\n\n

$=\\frac{2}{15} \\cdot \\sqrt{x-1}\\left[3 x^{2}+3-6 x+15+10 x-10\\right]+c$<\/p>\n\n\n\n

$=\\frac{2}{15} \\cdot \\sqrt{x-1} \\cdot\\left(3 x^{2}+4 x+8\\right)+c$<\/p>\n\n\n\n

Question 19<\/strong>
$\\int \\frac{x^{2}}{\\sqrt{1-x}} \\cdot d x$
Sol :
Let z=1-x then $\\frac{d z}{d x}=-1 \\quad \\therefore d x=-d z$<\/p>\n\n\n\n

Again z=1-x $\\therefore x=z+1$<\/p>\n\n\n\n

Now , $\\int \\frac{x^{2}}{\\sqrt{1-x}} d x$<\/p>\n\n\n\n

$=-\\int \\frac{(z+1)^{2}}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\int \\frac{(-z-1)^{2}}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\int \\frac{z^{2}+1+2 z}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\int \\frac{z^{z}}{\\sqrt{z}} d z+\\int \\frac{1}{\\sqrt{z}} d z+2 \\int \\frac{z}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\int z^{\\frac{3}{2}} d z+\\int z^{-\\frac{1}{2}} d z+2 \\int z^{\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=\\frac{2}{5} \\cdot 2^{\\frac{5}{2}}+2 \\cdot z^{\\frac{1}{2}}+2 \\cdot \\frac{2}{3} z^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

$=2z^{\\frac{1}{2}}\\left[\\frac{z^{2}}{5}+1+\\frac{2}{3} z\\right]+c$<\/p>\n\n\n\n

$=2 \\sqrt{z}\\left[\\frac{3z^{2}+15+10z}{15}\\right]+c$<\/p>\n\n\n\n

$=\\frac{2}{15} \\sqrt{z}\\left[3(x-1)^{2}+15+10(x-1)\\right]+c$<\/p>\n\n\n\n

$=\\frac{2}{15} \\sqrt{2}\\left[3\\left(x^{2}+1-2 x\\right)+15+10 x-10\\right]+c$<\/p>\n\n\n\n

$=\\frac{2}{15} \\sqrt{2}\\left[3 x^{2}+3-6 x+15+10 x-10\\right]+c$<\/p>\n\n\n\n

$=\\frac{2}{15} \\cdot \\sqrt{1-x}\\left[3 x^{2}+8+4 x\\right]+c$<\/p>\n\n\n\n

Question 20<\/strong>
$\\int \\frac{x}{\\sqrt{x+4}} d x \\quad x>0$
Sol :
Let z=x+4 then $\\frac{d z}{d x}=1 \\quad \\therefore d z=d x$<\/p>\n\n\n\n

Again z=x+4 $\\therefore x=z-4$<\/p>\n\n\n\n

Now , $\\int \\frac{x}{\\sqrt{x+4}} d x$<\/p>\n\n\n\n

$=\\int \\frac{z-4}{\\sqrt{z}} \\cdot d z$<\/p>\n\n\n\n

$=\\int \\frac{z}{\\sqrt{z}} d z-4 \\int \\frac{1}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\int \\sqrt{z} d z-4 \\int z^{-\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=\\int z^{\\frac{1}{2}} d z-4 \\int z^{-\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=\\frac{2}{3} \\cdot 2^{\\frac{3}{2}}-4 \\cdot 2 \\cdot z^{\\frac{1}{2}}+c$<\/p>\n\n\n\n

$=2z^{\\frac{1}{2}}\\left[\\frac{z}{3}-4\\right]+c$<\/p>\n\n\n\n

$=2 \\cdot \\sqrt{x+1}\\left[\\frac{x+1}{3}-4\\right]+c$<\/p>\n\n\n\n

$=2 \\sqrt{x+4}\\left(\\frac{x+4-12}{3}\\right)+c$<\/p>\n\n\n\n

$=\\frac{2}{3} \\cdot \\sqrt{x+4} \\cdot(x-8)+c$<\/p>\n\n\n\n

Question 21<\/strong>
$\\int \\frac{x^{2}}{\\sqrt{3 x+4}} d x$
Sol :
Let z=3x+4 then $\\frac{d z}{d x}=3 \\therefore d x=\\frac{d z}{3}$<\/p>\n\n\n\n

Again , z=3x+4 $\\therefore x=\\frac{z-4}{2}$<\/p>\n\n\n\n

Now , $\\int \\frac{x^{2}}{\\sqrt{3 x+1}} d x$<\/p>\n\n\n\n

$=\\int \\frac{\\left(\\frac{z-4}{3}\\right)^{2}}{\\sqrt{z}} \\frac{d z}{3}$<\/p>\n\n\n\n

$=\\int \\frac{(z-4)^{2}}{9 \\cdot \\sqrt{z}} \\cdot \\frac{d z}{3}$<\/p>\n\n\n\n

$=\\frac{1}{27} \\int \\frac{z^{2}+16-82}{\\sqrt{z}} d 2$<\/p>\n\n\n\n

$=\\frac{1}{27} \\int \\frac{z^{2}}{\\sqrt{z}} d z+\\frac{1}{27} \\cdot 16 \\int \\frac{1}{\\sqrt{z}} d z \\frac{-8}{27} \\int \\frac{z d}{\\sqrt{z}}$<\/p>\n\n\n\n

$=\\frac{1}{27} \\int z^{\\frac{3}{2}} d z+\\frac{16}{27} \\int z^{-\\frac{1}{2}} d z-\\frac{8}{27} \\int z^{\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=\\frac{1}{27} \\cdot \\frac{2}{5} \\cdot z^{\\frac{5}{2}}+\\frac{16}{27} \\cdot 2z^{\\frac{1}{2}}-\\frac{8\\times2}{27 \\times 3} z^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

$=\\frac{2}{27} z^{\\frac{1}{8}}\\left[\\frac{z^{2}}{5}+16-\\frac{8}{3} z\\right]+c$<\/p>\n\n\n\n

$=\\frac{2}{27} \\sqrt{3 x+4}\\left[\\frac{(3 x+4)^{2}}{5}-\\frac{8}{3}(3 x+4)+16\\right]+c$<\/p>\n\n\n\n

Question 22<\/strong>
$\\int \\frac{x^{2}}{\\sqrt{x+1}} d x$
Sol :
 Let z=x+1 then $\\frac{d z}{d x}=1 \\quad \\therefore d x=d 2$<\/p>\n\n\n\n

Again z=x+1 \u2234x=z-1<\/p>\n\n\n\n

Now , $\\int \\frac{x^{2}}{\\sqrt{x+1}} d x$<\/p>\n\n\n\n

$=\\int \\frac{(z-1)^{2}}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\int \\frac{z^{2}+1-2 z}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\int \\frac{z^{2}}{\\sqrt{z}} d z+\\int \\frac{1}{\\sqrt{z}} d z-2 \\int \\frac{z}{\\sqrt{z}} d z$<\/p>\n\n\n\n

$=\\int z^{\\frac{3}{2}} d z+\\int z^{-\\frac{1}{2}} d z-2 \\int z^{\\frac{1}{2}} d z$<\/p>\n\n\n\n

$=\\frac{2}{5} \\cdot z^{\\frac{5}{2}}+2 \\cdot z^{\\frac{1}{2}}-2 \\cdot \\frac{2}{3} \\cdot z^{\\frac{3}{2}}+c$<\/p>\n\n\n\n

$=2 z^{\\frac{1}{2}}\\left[\\frac{z^{2}}{5}+1-\\frac{2}{3} z\\right]+c$<\/p>\n\n\n\n

$=2 \\sqrt{x+1}\\left[\\frac{(x+1)^{2}}{5}+1-\\frac{2(x+1)}{3}\\right]+c$<\/p>\n\n\n\n

Question 23<\/strong>
$\\int \\frac{2 x-1}{(x-1)^{2}} d x$
Sol :
Let z=x-1 then $\\frac{d z}{d x}=1 \\quad \\therefore d z=d x$<\/p>\n\n\n\n

Again z=x-1 $\\therefore \\quad x=z+1$<\/p>\n\n\n\n

Now , $\\int \\frac{2 x+1}{(x-1)^{2}} d x$<\/p>\n\n\n\n

$=\\int \\frac{2(z+1)-1}{z^{2}} d z$<\/p>\n\n\n\n

$=\\int \\frac{2z+2-1}{z^{2}} d z$<\/p>\n\n\n\n

$=\\int \\frac{2 z+1}{z^{2}} d z$<\/p>\n\n\n\n

$=2 \\int \\frac{z}{z^{2}} d z+1 \\int \\frac{1}{z^{2}} d z$<\/p>\n\n\n\n

$=2 \\int \\frac{1}{z} d z+\\int z^{-2} d z$<\/p>\n\n\n\n

$=2 \\cdot \\log |z|+\\frac{z^{-2+1}}{-2+1}+c$<\/p>\n\n\n\n

$=2 \\log |x-1|-\\frac{1}{2}+c$<\/p>\n\n\n\n

$=2 \\log |x-1|-\\frac{1}{x-1}+c$<\/p>\n\n\n\n

Question 24<\/strong>
$\\int \\frac{x^{2}+3 x+2}{(x-2)(x+1)} d x$
Sol :
$=\\int \\frac{x^{2}+x+2 x+2}{(x-2)(x+1)} d x$<\/p>\n\n\n\n

$=\\int \\frac{x(x+1)+2(x+1)}{(x-2)(x+1)} d x$<\/p>\n\n\n\n

$=\\int \\frac{(x+1)(x+2)}{(x-2)(x+1)} d x$<\/p>\n\n\n\n

$=\\int \\frac{x+2}{x-2} d x$<\/p>\n\n\n\n

$=\\int \\frac{(x-2)+4}{(x-2)} d x$<\/p>\n\n\n\n

$=\\int \\frac{x+2}{x-2} d x+4 \\int \\frac{1}{x-2} d x$<\/p>\n\n\n\n

$=\\int d x+4 \\int \\frac{1}{x-2} d x$<\/p>\n\n\n\n

=x+4log|x+2|+c<\/p>\n\n\n\n

Question 25<\/strong>
$\\int \\frac{x^{2}+3 x+1}{(x+1)^{2}} d x$
Sol :
Let z=x+1 then $\\frac{d z}{d x}=1 \\quad \\therefore \\quad d z=d x$<\/p>\n\n\n\n

Again z=x+1 $\\cdots x=z-1$<\/p>\n\n\n\n

Now , $\\int \\frac{x^{2}+3 x+1}{(x+1)^{2}} d x$<\/p>\n\n\n\n

$=\\int \\frac{(z-1)^{2}+3(z-1)+1}{2^{z}} d z$<\/p>\n\n\n\n

$=\\int \\frac{z^{2}+1-2z+3z-3+1}{z^{2}} d z$<\/p>\n\n\n\n

$=\\int \\frac{z^{2}+2-1}{z^{2}} d z$<\/p>\n\n\n\n

$=\\int \\frac{z^{2}}{z^{2}} d z+\\int \\frac{z}{z^{2}} d z-\\int \\frac{1}{z^{2}} d z$<\/p>\n\n\n\n

$=\\int d z+\\int \\frac{1}{z} dz-\\int z^{-2} d z$<\/p>\n\n\n\n

$=2+\\log |2|-\\frac{z^{-2+1}}{-2+1}+c$<\/p>\n\n\n\n

$=z+\\log |z|+z^{-1}$<\/p>\n\n\n\n

$=z+\\log |z|+\\frac{1}{z}$<\/p>\n\n\n\n

$=(x+1)+\\log |x+1|+\\frac{1}{x+1}$<\/p>\n\n\n\n

$=x+\\log |x+1|+\\frac{1}{x+1}+c$<\/p>\n\n\n\n

Question 26<\/strong>
$\\int \\frac{d x}{x^{\\frac{1}{2}}+x^{\\frac{1}{3}}}$
Sol :
Let $x=2^{6} \\therefore z=x^{\\frac{1}{6}} \\therefore \\quad d x=6z^{5} d z$<\/p>\n\n\n\n

Now , $\\int \\frac{d x}{x^{\\frac{1}{2}}+x^{\\frac{1}{3}}}$<\/p>\n\n\n\n

$=\\int \\frac{6 z^{5} d z}{\\left(z^{6}\\right)^{1\/2} t+\\left(z^{6}\\right)^{1 \/ 3}}$<\/p>\n\n\n\n

$=\\int \\frac{6z^{5} d z}{z^{3}+z^{2}}$<\/p>\n\n\n\n

$=6 \\int \\frac{z^{5}}{z^{3}+z^{2}} d z$<\/p>\n\n\n\n

$=6 \\int \\frac{z^{5}d z}{z^{2}(z+1)}$<\/p>\n\n\n\n

$=6 \\int \\frac{z^{3}}{z+1} d z$<\/p>\n\n\n\n

$=6 \\int \\frac{z^{3}-1+1}{z+1} d z$<\/p>\n\n\n\n

$=6 \\int \\frac{(z+1)\\left(z^{2}+1-2\\right)}{(z+1)} d z-6 \\int \\frac{1}{z+1} d z$<\/p>\n\n\n

[a3<\/sup>+b3<\/sup>=(a+b)(a2<\/sup>+b2<\/sup>-ab)]<\/p>\n\n\n\n

$=6 \\int z^{2} d z+6 \\int d z-6 \\int z d z-6 \\log |z+1|+c$<\/p>\n\n\n\n

$=6 \\cdot \\frac{z^{3}}{3}+6z-\\frac{6z^{2}}{2}-6 \\log |z+1|+c$<\/p>\n\n\n\n

$=2z^{3}+6z-3z^{2}-6 \\log |z+1|+c$<\/p>\n\n\n\n

$=2 \\cdot\\left(x^{\\frac{1}{6}}\\right)^{3}+6 \\cdot x^{\\frac{1}{6}}-3\\left(x^{\\frac{1}{6}}\\right)^{2}-6 \\log \\left|1+x^{\\frac{1}{6}}\\right|+c$<\/p>\n\n\n\n

$=2 x^{\\frac{1}{2}}+6 x^{\\frac{1}{6}}-3 x^{\\frac{1}{3}}-6 \\log \\left|1+x^{\\frac{1}{6}}\\right|+c$<\/p>\n\n\n\n

$=2 \\sqrt{x}-3 x^{\\frac{1}{3}}+6 x^{\\frac{1}{6}}-6 \\log \\left|1+x^{\\frac{1}{8}}\\right|+c$<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 12 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Exercise 19.3 Question 1$\\displaystyle\\int(x+2) \\sqrt{3 x+5} d x$Sol :Let $z=3 x+5$ then $\\dfrac{d z}{d x}=3 $ \u2234$d x=\\dfrac{d z}{3}$ Again z=3 x+5 \u2234 $x=\\dfrac{2-5}{3}$ Now, $\\displaystyle\\int(x+2) \\cdot \\sqrt{3 x+5} d x$ $=\\displaystyle\\int\\left(\\dfrac{z-5}{3}+2\\right) \\cdot \\sqrt{z}  \\dfrac{dz}{3}$ $=\\frac{1}{3} \\displaystyle\\int\\left(\\frac{z-5+6}{3}\\right) \\sqrt{z} d z$ $=\\frac{1}{9} \\displaystyle\\int(z+1) \\sqrt{z} d z$ $=\\dfrac{1}{9} \\displaystyle\\int z^{\\tfrac{3}{2}} dz+\\dfrac{1}{9} \\displaystyle\\int z^{\\tfrac{1}{2}} d z$ $=\\dfrac{1}{9} \\cdot […]<\/p>\n","protected":false},"author":302,"featured_media":626470,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[25],"tags":[],"boards":[],"class_list":["post-626471","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-12","entry"],"yoast_head":"\nKC Sinha: Exercise 19.3- Mathematics Solution Class 12 Chapter 19 Indefinite Integrals - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Exercise 19.3 Question 1$displaystyleint(x+2) sqrt{3 x+5} d x$Sol :Let $z=3 x+5$ then $dfrac{d z}{d x}=3 $ \u2234$d x=dfrac{d z}{3}$ Again z=3 x+5 \u2234\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, 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