Text Book Solutions for Integral Calculus I Exercise 2.3

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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 2.4

Text Book Solutions for Integral Calculus I Exercise 2.5

Text Book Solutions for Integral Calculus I Exercise 2.6

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)

Text Book Solutions for Integral Calculus I Exercise 2.3

Integrate the following with respect to x.

$1. \: e^{\:x\: log a}+e^{\:a\: log a}-e^{\:n\: log x}$

$Let \: I =\int \left (e^{\:x\: log a}+e^{\:a\: log a}-e^{\:n\: log x} \right )dx\\$
$=\: \int\left (e^{\: log a^{x}}+e^{\: log a^{a}}-e^{\: log x^{n}} \right )dx\\$
$=\int \left (a^{x}+a^{a}-x^{n} \right )dx \: \: \: \because e^{log}=log e=1\\$
$=\frac{1}{log a}a^{x}+ a^{a}x-\frac{x^{n+1}}{n+1}+c$

$2.\dfrac{ a^{x}-e^{\:x\: log b}}{ e^{\:x\: log a}\: b^{x}}$

$Let \:I \:= \int \dfrac{ a^{x}-e^{\:x\: log b}}{ e^{\:x\: log a}\: b^{x}}\\$
$=\int \dfrac{a^{x}-e^{log\: bx}}{e^{log ax}\: bx}\\$
$=\int \dfrac{a^{x}-b^{x}}{a^{x}\: b^{x}}\: \: \: \because e^{log}=1\\$
$=\int \dfrac{a^{x}}{a^{x}\: b^{x}}-\dfrac{b^{x}}{a^{x}\: b^{x}}\\$
$=\int\left ( \dfrac{1}{b^{x}}-\dfrac{1}{a^{x}} \right )dx \\$
$=\int \left (b^{-x}-a^{-x} \right ) dx\\$
$=\frac{1}{log b}\: \dfrac{b^{-x}}{-1}-\dfrac{1}{log a}\: \dfrac{a^{-x}}{-1}+c\\$
$=-\dfrac{1}{a^{x}\: log a}+\dfrac{1}{b^{x}\: log b}+c\\$
$=\dfrac{1}{a^{x}\: log a}-\dfrac{1}{b^{x}\: log b}$

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

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

$4. \: \frac{e^{3x}-e^{-3x}}{e^{x}}$

$Let \: I \: = \int \left (\dfrac{e^{3x}-e^{-3x}}{e^{x}} \right )dx\\$
$=\int \left (\dfrac{e^{3x}}{e^{x}}-\dfrac{e^{-3x}}{e^{x}} \right )dx \\$
$=\int\left ( e^{3x}e^{-x}-e^{-3x}e^{-x} \right )dx\\$
$=\int \left (e^{2x}-e^{-4x} \right )dx\\$
$=\dfrac{e^{2x}}{2}-\dfrac{e^{-4x}}{-4}+c\\$
$=\dfrac{e^{2x}}{2}+\dfrac{e^{-4x}}{4}+c$

$5.\: \dfrac{e^{3x}+e^{5x}}{e^{x}+e^{-x}}$

$Let \: I \: = \int \left ( \dfrac{e^{3x}+e^{5x}}{e^{x}+e^{-x}} \right )dx \\$
$=\left (\dfrac{e^{3x}(1+e^{2x})}{e^{x}+\dfrac{1}{e^{x}}} \right )dx \\$
$=\int \left (\dfrac{e^{3x}(1+e^{2x})}{\dfrac{e^{2x}+1}{e^{x}}} \right )dx\\$
$=\int \left (\dfrac{e^{3x}e^{x}(1+e^{2x})}{e^{2x}+1} \right )dx\\$
$=\int \dfrac{e^{4x}}{4}dx \\$
$= \dfrac{e^{4x}}{4}+c$

$6.\: \left ( 1-\dfrac{1}{x^{2}} \right )e^{x+\dfrac{1}{x}}$

$Let \: I \: = \: \left ( 1-\dfrac{1}{x^{2}} \right )\left (e^{x+\dfrac{1}{x}} \right ) \:dx \\$
$We \: know \: that \: \int a\: e^{ax+b} d\left ( ax+b \right ) =\: e^{ax+b}+c\\$
$\therefore \: I = e^{\left (x+\dfrac{1}{x} \right )}+c$

$7. \: \frac{1}{x(logx)^{2}}$

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

$8. \: If\: f'(x)= e^{x} and \: f (0 )= 2,\: then\: find \:f(x )$

$\int f'(x)\: dx =\int e^{x}\: dx\\$
$=f(x)=e^{x}+c\cdots \cdots (1)\\$
$At \:x=0\\$
$f(0)=e^{0}+c\\$
$2=+\1+C\\$
$C=2-1\\$
$C=1\\$
$Substituting \: in \: Eq\: (1) \\$
$f(x)=e^{x}+1\\$