{"id":626488,"date":"2023-09-12T03:07:40","date_gmt":"2023-09-12T03:07:40","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=626488"},"modified":"2023-09-12T03:07:47","modified_gmt":"2023-09-12T03:07:47","slug":"kc-sinha-exercise-19-11-mathematics-solution-class-12-chapter-19-indefinite-integrals","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-19-11-mathematics-solution-class-12-chapter-19-indefinite-integrals\/","title":{"rendered":"KC Sinha: Exercise 19.11- Mathematics Solution Class 12 Chapter 19 Indefinite Integrals"},"content":{"rendered":"\n

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

Question 1<\/h4>\n\n\n\n

$\\int e^{x}\\left(\\frac{1+x \\log x}{x}\\right) d x$
Sol :
$I=\\int e^{x}\\left(\\frac{1+x \\log x}{x}\\right) d x=\\int e^{x}\\left(\\frac{1}{x}+\\frac{x \\log x}{x}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(\\frac{1}{x}+\\log x\\right) d x$<\/p>\n\n\n\n

$=\\int e^{x}\\left(\\log x+\\frac{1}{x}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where F(x)=logx<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x} \\log x+c$<\/p>\n\n\n\n

Question 2<\/h4>\n\n\n\n

$\\int e^{x}\\left(\\tan ^{-1} x+\\frac{1}{1+x^{2}}\\right) d x$
Sol :
$I=\\int e^{x}\\left(\\tan ^{-1} x+\\frac{1}{1+x^{2}}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where $F(x)=\\tan ^{-1} x$<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x} \\tan ^{-1} x+c$<\/p>\n\n\n\n

Question 3<\/h4>\n\n\n\n

$\\int e^{x}(\\sin x+\\cos x) d x$
Sol :
$I=\\int e^{x}(\\sin x+\\cos x) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where F(x)=sinx<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x} \\sin x+c$<\/p>\n\n\n\n

Question 4<\/h4>\n\n\n\n

$\\int e^{x}\\left(\\frac{1-\\sin x}{1-\\cos x}\\right) d x$
Sol :
$I=\\int e^{x}\\left(\\frac{1-\\sin x}{1-\\cos x}\\right) d x$<\/p>\n\n\n\n

$=\\int e^{x}\\left(\\frac{1-2 \\sin x \\cdot(\\cos x)}{2 \\sin ^{2} \\frac{x}{2}}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(\\frac{1}{2 \\sin ^{2} \\frac{x}{2}}-\\frac{2 \\frac{\\sin x}{2} \\cdot \\cos \\frac{x}{2}}{2 \\sin \\frac{x}{2}}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(\\frac{1}{2} \\operatorname{cosec}^{2} \\frac{x}{2}-\\frac{\\cot x}{2}\\right) d x$<\/p>\n\n\n\n

$=\\int e^{x}\\left(-\\cot \\frac{x}{2}+\\frac{1}{2} cosec ^{2} x\\right)dx$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(F(x)+F^{\\prime}(x)\\right) d x$<\/p>\n\n\n\n

where $F(x)=-\\cot \\frac{x}{2}$<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x}\\left(-\\cot \\frac{x}{2}\\right)+c$<\/p>\n\n\n\n

$=-e^{x} \\cot \\frac{x}{2}+c$<\/p>\n\n\n\n

Question 5<\/h4>\n\n\n\n

$\\int e^{x}\\left(\\frac{1+x}{(2+x)^{2}}\\right) d x$
Sol :
$I=\\int e^{x} \\frac{(1+x)}{(2+x)^{2}} d x=\\int e^{x} \\frac{(2+x-1)}{(2+x)^{2}} d x$<\/p>\n\n\n\n

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

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

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where F(x)=\\frac{1}{2+x}<\/p>\n\n\n\n

I=e^{x} F(x)+c=\\frac{e^{x}}{2+x}+c<\/p>\n\n\n\n

Question 6<\/h4>\n\n\n\n

$\\int e^{x} \\frac{(1-x)^{2}}{\\left(1+x^{2}\\right)^{2}} d x$
Sol :
$I=\\int e^{x} \\frac{(1-x)^{2}}{\\left(1+x^{2}\\right)^{2}} d x$<\/p>\n\n\n\n

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

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

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where $F(x)=\\frac{1}{1+x^{2}}$<\/p>\n\n\n\n

$I=e^{x} F(x)+c$<\/p>\n\n\n\n

$=e^{x} \\times \\frac{1}{1+x^{2}}+c=\\frac{e^{x}}{1+x^{2}}+c$<\/p>\n\n\n\n

Question 7<\/h4>\n\n\n\n

$e^{x} \\frac{\\left(x^{3}+x+1\\right)}{\\left(1+x^{2}\\right)^{3 \/ 2}} d x$
Sol :
$I=\\int e^{x} \\frac{\\left(x^{3}+x+1\\right)}{\\left(1+x^{2}\\right)^{3 \/ 2}} d x$<\/p>\n\n\n\n

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

$I=\\int e^{x}\\left(\\frac{x\\left(1+x^{2}\\right)}{\\left(1+x^{2}\\right)^{3 \/ 2}}+\\frac{1}{\\left(1+x^{2}\\right)^{-3\/ 2}}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(\\frac{x}{\\left(1+x^{2}\\right)^{\\frac{1}{2}}}+\\frac{1}{\\left(1+x^{2}\\right)^{3 \/ 2}}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$\\<\/p>\n\n\n\n

where $F(x)=\\frac{x}{\\left(1+x^{2}\\right)^{\\frac{1}{2}}} d x$<\/p>\n\n\n\n

$I=e^{x} F(x)+c$<\/p>\n\n\n\n

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

$\\frac{=x e^{x}}{\\sqrt{1+x^{2}}}+c$<\/p>\n\n\n\n

Question 8<\/h4>\n\n\n\n

(i)<\/strong> $\\int e^{x}(\\tan x-\\log \\cos x) d x$
Sol :
$I \\cdot \\int e^{x}(\\tan x-\\log \\cos x) d x$<\/p>\n\n\n\n

$I=\\int e^{x}(-\\log \\cos x+\\tan x) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(F(x)+F^{\\prime}(x)\\right) d x $<\/p>\n\n\n\n

where $F(x)=-\\log \\cos x$<\/p>\n\n\n\n

$I=e^{x} F(x)+c=-e^{x} \\log \\cos x+c$<\/p>\n\n\n\n

(ii)<\/strong> $\\int e^{x}\\left(\\cot x-\\operatorname{cosec}^{2} x\\right) d x$
Sol :
$I=\\int e^{x}\\left(\\cot x-\\operatorname{cosec}^{2} x\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[\\cot x +\\left(-cosec ^{2} x\\right)\\right] d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where F(x)=cotx<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x} \\cot x+c$<\/p>\n\n\n\n

(iii)<\/strong> $\\int e^{x}(\\tan x+\\log \\sec x) d x$
Sol :
$I=\\int e^{x}(\\tan x+\\log \\sec x) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where<\/p>\n\n\n\n

F(x)=logsecx<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x} \\cdot x \\log \\sec x+c$<\/p>\n\n\n\n

(iv)<\/strong> $\\int e^{x}\\left(\\frac{1+\\sin x \\cdot \\cos x}{\\cos ^{2} x}\\right) d x$
Sol :<\/p>\n\n\n\n

$I=\\int e^{x}\\left(\\frac{1+\\sin x \\cdot \\cos x}{\\cos ^{2} x}\\right) d x$<\/p>\n\n\n\n

$=\\int e^{x}\\left(\\frac{1}{\\cos ^{2} x}+\\frac{\\sin x \\cdot \\cos x}{\\cos ^{2} x}\\right)dx$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(\\sec ^{2} x+\\tan x\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where F(x)=tanx<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x} \\tan x+c$<\/p>\n\n\n\n

(v)<\/strong> $\\int e^{x}\\left(\\frac{\\sin x \\cdot \\cos x-1}{\\sin ^{2} x}\\right) d x$
Sol :
$I=\\int e^{x}\\left(\\frac{\\sin x \\cdot \\cos x}{\\sin ^{2} x}-\\frac{1}{\\sin ^{2} x}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(\\cot x-\\operatorname{cosec}^{2} x\\right) d x$<\/p>\n\n\n\n

$I=e^{x} \\cot x+c$<\/p>\n\n\n\n

(vi)<\/strong> $\\int e^{x}(\\cot x+\\log \\sin x) d x$
Sol :
$I=\\int e^{x}(\\cot x+\\log \\sin x) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where F(x)=logsinx<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x} \\log \\sin x+c$<\/p>\n\n\n\n

(vii)<\/strong> $\\int e^{x} \\sec x(1+\\tan x) d x$
Sol :
$I=\\int e^{x} \\sec x(1+\\tan x) d x$<\/p>\n\n\n\n

$=\\int e^{x}(\\sec x+\\sec x \\cdot \\tan x) d x$<\/p>\n\n\n\n

$I=e^{x} \\sec x+c$<\/p>\n\n\n\n

Question 9<\/h4>\n\n\n\n

$\\int e^{x} \\frac{\\left[1+\\sqrt{1-x^{2}} \\sin ^{-1} x\\right]}{\\sqrt{1-x^{2}}} d x$
Sol :
$I=\\int e^{x} \\frac{\\left[1+\\sqrt{1-x^{2}} \\sin ^{-1} x\\right]}{\\sqrt{1-x^{2}}} d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(\\frac{1}{\\sqrt{1-x^{2}}}+\\frac{\\sqrt{1-x^{2}} \\cdot \\sin ^{-1} x}{\\sqrt{1-x^{2}}}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(\\sin ^{-1} x+\\frac{1}{\\sqrt{1-x^{2}}}\\right) d x$<\/p>\n\n\n\n

$I=e^{x} \\sin ^{-1} x+c$<\/p>\n\n\n\n

Question 10<\/h4>\n\n\n\n

(i)<\/strong> $\\int e^{x}\\left(\\frac{1}{x^{2}}-\\frac{2}{x^{3}}\\right) d x$
Sol :
$I=\\int e^{x}\\left(\\frac{1}{x^{2}}-\\frac{2}{x^{3}}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where $F(x)=\\frac{1}{x^{2}}$<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x} \\times \\frac{1}{x^{2}}+c$<\/p>\n\n\n\n

$=\\frac{e^{x}}{x^{2}}+c$<\/p>\n\n\n\n

(ii)<\/strong> $\\int e^{x}\\left(\\frac{1}{x}-\\frac{1}{x^{2}}\\right) d x$
Sol :
$I=\\int e^{x}\\left(\\frac{1}{x}-\\frac{1}{x^{2}}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where $F(x)=\\frac{1}{x}$<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x} x \\frac{1}{x}+c$<\/p>\n\n\n\n

$=\\frac{e^{x}}{x}+c$<\/p>\n\n\n\n

(iii)<\/strong> $\\int e^{2x}\\left(\\frac{2 x-1}{4 x^{2}}\\right) d x$
Sol :
$I=\\int e^{2 x}\\left(\\frac{2 x-1}{4 x^{2}}\\right) d x$<\/p>\n\n\n\n

putting z=2x , $z^{2}=4x^{2}$then dz=2dx $\\Rightarrow \\frac{d z}{2}=d x$<\/p>\n\n\n\n

Now , $I=\\int e^{2}\\left(\\frac{z-1}{z^{2}}\\right) \\frac{d z}{2}$<\/p>\n\n\n\n

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

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

$I=\\frac{1}{2} \\cdot e^{z} \\times \\frac{1}{2}+c$<\/p>\n\n\n\n

$=\\frac{1}{2} \\frac{e^{2 x}}{2 x}$<\/p>\n\n\n\n

$=\\frac{e^{2 x}}{4 x}+c$<\/p>\n\n\n\n

(iv)<\/strong> $\\int \\frac{x-3}{(x+1)^{3}} \\cdot e^{x} d x$
Sol :
$I=\\int \\frac{(x-3)}{(x-1)^{3}} e^{x} d x$<\/p>\n\n\n\n

$=\\int e^{x}\\left(\\frac{x-3}{(x-1)^{3}}\\right) d x$<\/p>\n\n\n\n

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

$=\\int e^{x}\\left[\\frac{(x-1)}{(x-1)^{2}}-\\frac{2}{(x-1)^{3}}\\right] d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(\\frac{1}{(x-1)^{2}}-\\frac{2}{(x-1)^{3}}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where $F(x)=\\frac{1}{(x-1)^{2}}$<\/p>\n\n\n\n

$I=e^{x} F(x)+c$<\/p>\n\n\n\n

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

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

(v)<\/strong> $\\int \\frac{(2-x) e^{x}}{(1-x)^{2}} d x$
Sol :
$I=\\int \\frac{2-x \\cdot e^{x}}{(1-x)^{2}} d x$<\/p>\n\n\n\n

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

$I=\\int\\left(\\frac{(1-x)}{(1-x)^{2}}+\\frac{1}{(1-x)^{2}}\\right) \\cdot e^{x} d x$<\/p>\n\n\n\n

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

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where $F(x)=\\frac{1}{1-x}$<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x} \\frac{1}{1-x}+c$<\/p>\n\n\n\n

$=\\frac{e^{x}}{1-x}+c$<\/p>\n\n\n\n

Question 11<\/h4>\n\n\n\n

$\\int e^{x}\\left(\\frac{2-\\sin 2 x}{1-\\cos 2 x}\\right) d x$
Sol :
$I=\\int e^{x}\\left(\\frac{2-2 \\sin x \\cdot \\cos x}{2 \\sin ^{2} x}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(\\frac{2}{2 \\sin ^{2} x}-\\frac{2 \\sin x \\cdot \\cos x}{2 \\sin ^{2} x}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(\\cos e c^{2} x-\\cot x\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(-\\cot x+\\operatorname{cosec}^{2} x\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where F(x)=-cotx<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x}(-\\cot x)+c$<\/p>\n\n\n\n

$=-e^{x} \\cot x+c$<\/p>\n\n\n\n

Question 12<\/h4>\n\n\n\n

$\\int e^{x}\\left(\\frac{2+\\sin 2 x}{1+\\cos 2 x}\\right) d x$
Sol :
$I=\\int e^{x}\\left(\\frac{2+\\sin 2 x}{1+\\cos 2 x}\\right) d x$<\/p>\n\n\n\n

$=\\int e^{x}\\left(\\frac{2+2 \\sin x \\cdot \\cos x}{2 \\cos ^{2} x}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left(\\frac{2}{2 \\cos ^{2} x}+\\frac{2 \\sin x \\cdot \\cos x}{2 \\cos ^{2} x}\\right) d x$<\/p>\n\n\n\n

$=\\int e^{x}\\left(\\sec ^{2} + \\tan x\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where F(x)=tanx<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x} \\tan x+c$<\/p>\n\n\n\n

Question 13<\/h4>\n\n\n\n

$\\int e^{2 x}(-\\sin x+2 \\cos x) d x$
Sol :
putting 2x=z then dz=2dx $\\Rightarrow \\frac{d z}{2}=d x$<\/p>\n\n\n\n

Now , $I=\\int e^{z}\\left(-\\sin \\frac{z}{2}+2 \\cos \\frac{z}{2}\\right) \\frac{d z}{2}$<\/p>\n\n\n\n

$I=\\frac{1}{2} \\int e^{2}\\left(2 \\cos \\frac{z}{2}-\\sin \\frac{z}{2}\\right) d z$<\/p>\n\n\n\n

$I=\\int e^{2}\\left(\\cos \\frac{z}{2}-\\frac{1}{2} \\sin \\frac{z}{2}\\right) d z$<\/p>\n\n\n\n

$I=\\int e^{z}\\left[F(z)+F^{\\prime}(z)\\right] d z$<\/p>\n\n\n\n

where $F(z)=\\cos \\frac{z}{2}$<\/p>\n\n\n\n

$I=e^{z} F(z)+c$<\/p>\n\n\n\n

$=\\frac{e^{2} \\cdot \\cos z}{2 }+c$<\/p>\n\n\n\n

$=e^{2 x} \\cdot \\cos \\frac{2 x}{2}+c$<\/p>\n\n\n\n

$=e^{2 x} \\cos x+c$<\/p>\n\n\n\n

Question 14<\/h4>\n\n\n\n

(i)<\/strong> $\\left[\\tan (\\log x)+\\sec ^{2}(\\log x)\\right] d x$
Sol :
putting z=logx  $\\Rightarrow e^{2}=x \\Rightarrow d x=e^{2} d z$<\/p>\n\n\n\n

Now , $I=\\int\\left(\\tan z+\\sec ^{2} z\\right) \\cdot e^{z} d z$<\/p>\n\n\n\n

$I=\\int e^{2}\\left[F(z)+F^{\\prime}(z)\\right] d z$<\/p>\n\n\n\n

where F(z)=tanz<\/p>\n\n\n\n

$I=e^{z} F(z)+c=e^{2} \\tan z+c$<\/p>\n\n\n\n

$=e^{\\log x} \\tan (\\log x)+c$ $\\left(\\because e^{\\log x}=x\\right)$<\/p>\n\n\n\n

=x.tan(log x)+c<\/p>\n\n\n\n

(ii)<\/strong> $\\int e^{x}[\\sec x+\\log (\\sec x+\\tan x)] d x$
Sol :
$I=\\int e^{x} \\left[ \\sec x+\\log (\\sec x+\\tan x)\\right] d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where F(x)=log(secx +tanx)<\/p>\n\n\n\n

$I=e^{x} F(x)+c$<\/p>\n\n\n\n

$=e^{x} \\log (\\sec x+\\tan x)+c$<\/p>\n\n\n\n

Question 15<\/h4>\n\n\n\n

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

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

$I=\\int e^{x} \\frac{\\left(1+x^{2}-2 x-4+4\\right)}{(1+x)^{2}} d x$<\/p>\n\n\n\n

$=\\int e^{x}\\left(\\frac{x^{2}-2 x-3+4}{(1+x)^{2}}\\right) d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[\\frac{x^{2}-2 x-3}{(1+x)^{2}}+\\frac{4}{(1+x)^{2}}\\right] d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[\\frac{x^{2}-3 x+x-3}{(1+x)^{2}}+\\frac{4}{(1+x)^{2}}\\right] d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[\\frac{x(x-3)+(x-3)}{(1+x)^{2}}+\\frac{4}{(1+x)^{2}}\\right] d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[\\frac{(x-3)(x+1)}{(1+x)^{2}}+\\frac{4}{(1+x)^{2}}\\right] d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[\\left(\\frac{x-3}{1+x}\\right)+\\frac{4}{(1+x)^{2}}\\right] d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where $F(x)=\\left(\\frac{x-3}{1+x}\\right)$<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x} \\times \\frac{(x-3)}{1+x}+c$<\/p>\n\n\n\n

(ii)<\/strong> $\\int \\frac{e^{x}\\left(x^{3}-x+2\\right)}{\\left(1+x^{2}\\right)^{2}} d x$
Sol :
$I=\\int e^{x} \\frac{\\left(x^{3}-x+2\\right)}{\\left(1+x^{2}\\right)^{2}} d x$<\/p>\n\n\n\n

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

$I=\\int e^{x}\\left[\\frac{\\left(1+x^{2}\\right)(1+x)}{\\left(1+x^{2}\\right)^{2}}+\\frac{1-2 x-x^{2}}{\\left(1+x^{2}\\right)^{2}}\\right] d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[\\left(\\frac{1+x}{1+x^{2}}\\right)+\\left(\\frac{1-2 x-x^{2}}{\\left(1+x^{2}\\right)^{2}}\\right)\\right] d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where $F(x)=\\frac{1+x}{1+x^{2}}$<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x}\\left(\\frac{1+x}{x^{2}+1}\\right)+c$<\/p>\n\n\n\n

(iii)<\/strong> $\\int \\frac{x \\cdot e^{2 x}}{(1+2 x)^{2}} d x$
Sol :
putting z=2x then dz=2dx $\\Rightarrow d x=\\frac{d z}{2}$<\/p>\n\n\n\n

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

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

$=\\frac{1}{4} \\int e^{2}\\left(\\frac{1+z-1}{(1+z)^{2}}\\right) d z$<\/p>\n\n\n\n

$I=\\frac{1}{4} \\int e^{2}\\left(\\frac{1+z}{(1+z)^{2}}-\\frac{1}{(1+z)^{2}}\\right) d z$<\/p>\n\n\n\n

$=\\frac{1}{4} \\int e^{2}\\left(\\frac{1}{1+z}-\\frac{1}{(1+z)^{2}}\\right) d z$<\/p>\n\n\n\n

$I=\\frac{1}{4} \\cdot e^{2} \\times \\frac{1}{1+z}+c$<\/p>\n\n\n\n

$=\\frac{1}{4} \\cdot \\frac{e^{2 x}}{1+2 x}+c$<\/p>\n\n\n\n

$=\\frac{e^{2 x}}{4(1+2 x)}+c$<\/p>\n\n\n\n

Question 16<\/h4>\n\n\n\n

$\\int e^{\\tan ^{2} x}\\left(1+\\frac{x}{1+x^{2}}\\right) d x$
Sol :
$I=\\int e^{\\tan ^{-1} x}\\left(1+\\frac{x}{1+x^{2}}\\right) d x$<\/p>\n\n\n\n

putting $z=\\tan ^{-1} x$
\u21d2x=tanz<\/p>\n\n\n\n

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

Now  , $I=\\int e^{\\tan ^{-1} x}\\left(\\frac{1+x^{2}+x}{1+x^{2}}\\right) d x$<\/p>\n\n\n\n

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

$I=\\int e^{z}\\left(\\sec ^{2} z+\\tan z\\right) d z$<\/p>\n\n\n\n

$=e^{z} \\tan z+c$<\/p>\n\n\n\n

$I=e^{\\tan ^{-1} x} \\tan \\tan ^{-1} x+c$<\/p>\n\n\n\n

$=x \\cdot e^{\\tan ^{-1} x}+c$<\/p>\n\n\n\n

Question 17<\/h4>\n\n\n\n

$\\left.\\int e^{x} \\frac{1}{x} \\int x(\\log x)^{2}+2 \\log x\\right] d x$
Sol :
$I=\\int e^{x} \\frac{1}{x}\\left[x(\\log x)^{2}+2 \\log x\\right] d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[(\\log x)^{2}+\\frac{2 \\log x}{x}\\right] d x$<\/p>\n\n\n\n

$I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$<\/p>\n\n\n\n

where $F(x)=(\\log x)^{2}$<\/p>\n\n\n\n

$I=e^{x} F(x)+c=e^{x}(\\log x)^{2}+c$<\/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":"

Question 1 $\\int e^{x}\\left(\\frac{1+x \\log x}{x}\\right) d x$Sol :$I=\\int e^{x}\\left(\\frac{1+x \\log x}{x}\\right) d x=\\int e^{x}\\left(\\frac{1}{x}+\\frac{x \\log x}{x}\\right) d x$ $I=\\int e^{x}\\left(\\frac{1}{x}+\\log x\\right) d x$ $=\\int e^{x}\\left(\\log x+\\frac{1}{x}\\right) d x$ $I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] d x$ where F(x)=logx $I=e^{x} F(x)+c=e^{x} \\log x+c$ Question 2 $\\int e^{x}\\left(\\tan ^{-1} x+\\frac{1}{1+x^{2}}\\right) d x$Sol :$I=\\int e^{x}\\left(\\tan ^{-1} x+\\frac{1}{1+x^{2}}\\right) d x$ $I=\\int e^{x}\\left[F(x)+F^{\\prime}(x)\\right] […]<\/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-626488","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-12","entry"],"yoast_head":"\nKC Sinha: Exercise 19.11- Mathematics Solution Class 12 Chapter 19 Indefinite Integrals - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1 $int e^{x}left(frac{1+x log x}{x}right) d x$Sol :$I=int e^{x}left(frac{1+x log x}{x}right) d x=int e^{x}left(frac{1}{x}+frac{x\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-19-11-mathematics-solution-class-12-chapter-19-indefinite-integrals\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 19.11- Mathematics Solution Class 12 Chapter 19 Indefinite Integrals\" \/>\n<meta property=\"og:description\" content=\"Question 1 $int e^{x}left(frac{1+x log x}{x}right) d x$Sol :$I=int e^{x}left(frac{1+x log x}{x}right) d x=int e^{x}left(frac{1}{x}+frac{x\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-19-11-mathematics-solution-class-12-chapter-19-indefinite-integrals\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" 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