Integral Calculus I Exercise 2.6 - 12th Text Book Solutions

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Important Formulas in Integral Calculus I

Text Book Solutions for Integral Calculus I Exercise 2.1

Text Book Solutions for Integral Calculus I Exercise 2.2

Text Book Solutions for Integral Calculus I Exercise 2.3

Text Book Solutions for Integral Calculus I Exercise 2.4

Text Book Solutions for Integral Calculus I Exercise 2.5

Text Book Solutions for Integral Calculus I Exercise 2.7

Text Book Solutions for Integral Calculus I Exercise 2.8

Text Book Solutions for Integral Calculus I Exercise 2.9

Text Book Solutions for Integral Calculus I Exercise 2.10

Text Book Solutions for Integral Calculus I Exercise 2.11

Text Book Solutions for Integral Calculus I Exercise 2.12 (MCQ)

Integral Calculus I Exercise 2.6 -Text Book Solutions

Integrate the following with respect to x

$1. \dfrac{2x+5}{x^{2}+5x+7}$

$Let \: I =\int \dfrac{2x+5}{x^{2}+5x+7} dx\\$
$Put \: t=x^{2}+5x+7\\$
$dt = 2x+5 dx\\$
$I =\int \frac{dt}{t}\\$
$=log \left | t \right |+c\\$
$=log \left | x^{2}+5x+7 \right |+c$

$2. \dfrac{e^{3 logx}}{x^{4}+1}}$

$I =\int \dfrac{e^{3 logx}}{x^{4}+1}dx\\$
$=\int \dfrac{e^{logx^{3}}}{x^{4}+1}dx\\$
$=\frac{x^{3}}{{x^{4}+1}}dx\\$
$Put \: t= {x^{4}+1}dx\\$
$dt =4x^{3}dx\\$
$\frac{dt}{4}=x^{3}dx\\$
$=\int \frac{\frac{dt}{4}}{t}dt\\$
$=\frac{1}{4}\int \frac{dt}{t}\\$
$=\frac{1}{4} log\left | t \right |+c\\$
$=\frac{1}{4} log\left | x^{4}+1 \right |+c$

$3. \dfrac{e^{2x}}{e^{2x}-2}$

$Let \: I = \int \dfrac{e^{2x}}{e^{2x}-2}dx\\$
$Put \: t={e^{2x}-2} dx\\$
$dt =e^{2x}2 dt\\$
$\dfrac{dt}{2}=e^{2x}dx\\$
$I=\int \dfrac{\dfrac{dt}{2}}{t}dt\\$
$=\dfrac{1}{2}\int \dfrac{dt}{t}\\$
$=\dfrac{1}{2} log\left | t \right |+c\\$
$=\dfrac{1}{2} log\left |{e^{2x}-2} \right |+c\\$

$4.\: \dfrac{(logx)^{3}}{x}$

$Let\: I = \int \frac{(logx)^{3}}{x}dx\\$
$Put\: t=logx\\$
$dt=\dfrac{1}{x}dx\\$
$I=\int t^{3}dt\\$
$=\dfrac{t^{4}}{4}+c\\$
$=\frac{(logx)^{4}}{4}+c$

$5. \: \dfrac{6x+7}{\sqrt{3x^{2}+7x-1}}$

$Let \: I =\int \dfrac{6x+7}{\sqrt{3x^{2}+7x-1}}\\$
$t=3x^{2}+7x-1\\$
$dt=6x+7 \:dx\\$
$=\int \dfrac{dt}{\sqrt{t}}dt\\$
$=\int \dfrac{dt}{t^{\frac{1}{2}}}\\$
$=\int t^{\frac{-1}{2}}dt\\$
$=\dfrac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\\$
$=\dfrac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\\$
$=2t^{\frac{1}{2}}+c\\$
$=2\sqrt{t}+c\\$
$=2\sqrt{3x^{2}+7x-1}+c$

$6. \: (4x+2)\sqrt{x^{2}+x+1}$

$Let \: I = \int (4x+2)\sqrt{x^{2}+x+1}\:dx\\$
$=\int 2 (2x+1) (\sqrt{x^{2}+x+1})\: dx\\$
$2\int (2x+1) (\sqrt{x^{2}+x+1})\:dx\\$
$Put\: t=x^{2}+x+1\\$
$dt=(2x+1) dx\\$
$=2\int \sqrt{t}\:dt\\$
$=2\int t^{\frac{1}{2}}\:dt\\$
$=2 \left (\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\\ \right )+c\\$
$=2\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\\$
$=\frac{4}{3}t^{\frac{3}{2}}+c\\$
$=\frac{4}{3}(x^{2}+7x-1)^{\frac{3}{2}}+c$

$7. \: x^{8}(1+x^{9})^{5}$

$Let \: I = \int x^{8}(1+x^{9})^{5}\:dx\\$
$Put \: t=x^{9}\\$
$dt=9x^{8} \: dt\\$
$\frac{dt}{9}=x^{8}\:dx\\$
$I =\int (1+t)^{5} \: \frac{dt}{9}\\$
$=\frac{1}{9}\int (1+t)^{5}dt\\$
$=\dfrac{1}{9}\dfrac{(1+t)^{6}}{6}+c\\$
$=\dfrac{(1+t)^{6}}{54}+c\\$
$=\dfrac{(1+x^{9})^{6}}{54}+c$

$8. \: \dfrac{x^{e-1}+e^{x-1}}{x^{e}+e^{x}}$

$Let \: I= \int \dfrac{x^{e-1}+e^{x-1}}{x^{e}+e^{x}}\:dx\\$
$Put \: t=x^{e}+e^{x}\\$
$dt=(ex^{e-1}+e^{x})\: dx\\$
$dt = e(x^{e-1}+e^{x-1})\: dx\\$
$\frac{dt}{e} =(x^{e-1}+e^{x-1})\: dx\\$
$I=\int \frac{\frac{dt}{e}}{t}\\$
$=\dfrac{1}{e}\int \dfrac{dt}{t}\\$
$=\frac{1}{e} log\left | t \right |+c\\$
$=\frac{1}{e} log\left | x^{e}+e^{x} \right |+c$

$9.\: \dfrac{1}{xlogx}$

$Let \: I =\int \dfrac{1}{xlogx}\: dx\\$
$Put \: t=logx\\$
$dt=\frac{1}{x}\: dx\\$
$I=\int \frac{dt}{t}\\$
$=log\left | t \right |+c\\$
$=log\left | logx \right |+c$

$10.\: \dfrac{x}{2x^{4}-3x^{2}-2}$

$Let \: I = \int \dfrac{x}{2x^{4}-3x^{2}-2}\: dx\\$
$Put \: t=x^{2}\\$
$dt=2x\: dx\\$
$\dfrac{dt}{2}=x \: dx\\$
$I=\int \dfrac{\dfrac{dt}{2}}{2t^{2}-3t-2}\\$
$Factorising \: 2t^{2}-3t-2\\$
$=2t^{2}-4t+t-2 \\$
$2t(t-2)+1(t-2)\\$
$=(2t+1)(t-2)\\$
$I =\dfrac{1}{2}\int\dfrac{dt}{(t-2)(2t+1)}\\$
$=\dfrac{1}{(t-2)(2t+1)}=\dfrac{A}{(t-2)}+\dfrac{B}{(2t+1)}\\$
$=\dfrac{1}{(t-2)(2t+1)}= \dfrac{A(2t+1)+B(t-2)}{(t-2)(2t+1)}\\$
$1=A(2t+1)+B(t-2)\cdots (1)\\$
$Put\: t=2 \: in \: (1)\\$
$1=A(4+1)+0\\$
$1=5A\\$
$A=\dfrac{1}{5}\\$
$Put \: t= 0 \: in (1)\\$
$1= A(0+1)+B(0-2)\\$
$1=\dfrac{1}{5}(1)-2B\\$
$1-\dfrac{1}{5}=-2B\\$
$\dfrac{4}{5}=-2B\\$
$\dfrac{-2}{5}=B\\$
$=\dfrac{1}{(t-2)(2t+1)}= \dfrac{\dfrac{1}{5}}{(t-2)}-\dfrac{\dfrac{2}{5}}{(2t+1)}\\$
$I=\dfrac{1}{2}\int \dfrac{1}{5(t-2)}-\dfrac{2}{5(2t+1)}\: dt\\$
$\dfrac{1}{2} \left [ \dfrac{1}{5}log\left | t-2 \right |-\dfrac{2}{5}\: \dfrac{log\left | 2t+1 \right |}{2} \right ]+c\\$
$\dfrac{1}{2} \left [ \dfrac{1}{5}log\left | t-2 \right |-\dfrac{1}{5}\: log\left | 2t+1 \right | \right ]+c\\$
$\dfrac{1}{10}\left [ log\left | t-2 \right |-log\left | 2t+1 \right | \right ]+c\\$
$\dfrac{1}{10}log\left | \dfrac{t-2}{2t+1} \right |+c\\$
$\dfrac{1}{10}log\left | \dfrac{x^{2}-2}{2x^{2}+1} \right |+c$

$11. \: e^{x}(1+x)log xe^{x}$

$Let \: I = \int e^{x}(1+x)\: log x\: e^{x} \: dx\\$
$Put \: t=x\: e^{x}\\$
$dt=1(e^{x})+x(e^{x})\: dx\\$
$dt=e^{x}(1+x)\: dx\\$
$I =\int log t \: dt\\$
$Integrating \: by \: parts\\$
$u=log t\\$
$du=\frac{1}{t}\: dt\\$
$dv=dt\\$
$\int dv=\int dt\\$
$v=t\\$
$\int udv=uv-\int vdu\\$
$I= log t(t)-\int t(\frac{1}{t})\: dt\\$
$t \: log t -\int dt\\$
$t\: log t-t+c\\$
$t(logt-1)+c\\$
$=xe^{x}(log x\: e^{x}-1)+c$

$12. \: \dfrac{1}{x\: (x^{2}+1)}$

$Let \: I =\dfrac{1}{x\: (x^{2}+1)}\: dx\\$
$Using \: partial\: fractions\\$
$\dfrac{1}{x\: (x^{2}+1)}=\dfrac{A}{x}+\dfrac{Bx+C}{(x^{2}+1)}\\$
$\dfrac{1}{x\: (x^{2}+1)} = \dfrac{A(x^{2}+1)+(Bx+C)x}{(x^{2}+1)}\\$
$1= A(x^{2}+1)+(Bx+C)x\cdots(1)\\$
$Put \: x=0 \: in (1) \\$
$1=A+0\\$
$A=1\\$
$Put\: x=1 \: in (1) \\$
$1= A(1+1)+ (B(1)+C)(1)\\$
$1= 1(2)+B+C\\$
$1=2+B+C\\$
$1-2=B+C\\$
$-1= B+C \cdots (2)\\$
$Put \: x=2 \: in \cdots (1)\\$
$1=A(4+1)+(2B+C)(2)\\$
$1= 5A +4B+2C\\$
$1= 5(1)+4B+2C\\$
$1-5=4B+2C\\$
$-4=4B+2C\cdots (3)\\$
$Solving \: (2)\: and \: (3)\\$
$B+C=-1\\$
$2B+C=-2\\$
$\noindent\rule{2.5cm}{0.5pt}\\$
$-B=1\\$
$B=-1\\$
$Substituting \: B=-1 \: in \:(2) \\$
$-1= -1+C \\$
$C=0\\$
$\dfrac{1}{x(x^{2}+1)}=\dfrac{1}{x} +\dfrac{(-1)x+0}{x^{2}+1}\\$
$\dfrac{1}{x(x^{2}+1)}=\dfrac{1}{x}-\dfrac{x}{x^{2}+1}\\$
$I=\int \dfrac{1}{x(x^{2}+1)}=\dfrac{1}{x}-\dfrac{x}{x^{2}+1}\\$
$\int \dfrac{1}{x}\: dx-\int \dfrac{x}{x^{2}+1} \: dx\\$
$I =log \left | x \right |- I_{1}\\$
$Let\: I_{1} =\int \dfrac{x}{x^{2}+1} \: dx\\$
$Put \: t=x^{2}+1 \: dx\\$
$dt=2x \: dx\\$
$\dfrac{dt}{2}= x\: dx\\$
$I_{1}=\int \dfrac{\dfrac{dt}{2}}{t}\\$
$=\dfrac{1}{2}\int \dfrac{dt}{t}\\$
$=\frac{1}{2}log \left | t \right |+c\\$
$=\dfrac{1}{2}log \left | t \right |+c\\$
$=\dfrac{1}{2}log \left | x^{2} +1\right |+c\\$
$=log x-\dfrac{1}{2}log \left | x^{2} +1\right |+c$

$13. \: e^{x}\left ( \dfrac{1}{x^{2}} -\dfrac{2}{x^{3}}\right )$

$Let \: I = \int e^{x}\left ( \dfrac{1}{x^{2}} -\dfrac{2}{x^{3}}\right ) \: dx\\$
$Let f(x)=\dfrac{1}{x^{2}}\\$
$f^{'}(x) = \dfrac{-2}{x^{3}}\\$
$\int e^{x}\left ( f(x) +f^{'}(x)\right )\: dx = e^{x} f(x)+c\\$
$I =e^{x}\left (\frac{1}{x^{2}} \right )+c$

$14. \: e^{x}\left ( \dfrac{x-1}{(x+1)^{3}} \right )$

$Let \: I =\int e^{x}\left ( \dfrac{x-1}{(x+1)^{3}} \right )\: dx\\$
$=\int e^{x}\left ( \dfrac{x-1-1+1}{(x+1)^{3}} \right )\: dx \: (Adding \: and \: subtracting \: 1)\\$
$=\int e^{x}\left ( \dfrac{(x+1-2)}{(x+1)^{3}} \right )\: dx\\$
$=\int e^{x}\left (\dfrac{x+1}{(x+1)^{3}} -\dfrac{2}{(x+1)^{3}} \right )\: dx\\$
$=\int e^{x}\left (\dfrac{1}{(x+1)^{2}} -\dfrac{2}{(x+1)^{3}} \right )\: dx\\$
$Let \: f(x)= \dfrac{1}{(x+1)^{2}}\\$
$f^{'}(x) = \dfrac{-2}{(x+1)^{3}}\\$
$\int e^{x}\left ( f(x) +f^{'}(x)\right )\: dx = e^{x} f(x)+c\\$
$I = e^{x}\left ( \dfrac{1}{(x+1)^{2}} \right )+c$

$14. \: e^{3x}\left ( \dfrac{(3x-1)}{9x^{2}} \right )$

$Let\: I =\int e^{3x}\left ( \dfrac{(3x-1)}{9x^{2}} \right )\: dx\\$
$=e^{3x}\left ( \dfrac{3x}{9x^{2}} -\dfrac{1}{9x^{2}}\right )\: dx\\$
$=e^{3x}\left ( \dfrac{3}{9x} -\dfrac{1}{9x^{2}}\right )\: dx\\$
$Let \: f(x)=\dfrac{1}{9x}\\$
$f^{'}(x)=\dfrac{-1}{9x^{2}}\\$
$\int e^{ax}\left ( af(x) +f^{'}(x)\right )\: dx = e^{ax} f(x)+c\\$
$I =e^{3x}\left ( \dfrac{1}{9x} \right )+c\\$
$=\dfrac{e^{3x}}{9x}+c$

Integral Calculus I Exercise 2.6 – 12th Text Book Solutions (Business Maths)

Integral Calculus I Exercise 2.6 -Text Book Solutions

Integrate the following with respect to x

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Integral Calculus I Exercise 2.6 -Text Book Solutions

Integrate the following with respect to x

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