Text Book Solutions for Integral Calculus I Exercise 2.2

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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 2.3

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 2.6

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 Exercise 2.2

Integrate the following with respect to x.

$1. \:\left (\sqrt{2x}-\frac{1}{\sqrt{2x}} \right )^{2}$

$Let \: I\: = \int \:\left (\sqrt{2x}-\frac{1}{\sqrt{2x}} \right )^{2} dx$

$\int \left (2x-2+\frac{1}{2x} \right )dx$

$= \frac{2x^{2}}{2}-2x+\frac{1}{2}log \left | x \right |+c$

$= x^{2}-2x +\frac{1}{2}log \left | x \right |+c$

$2. \: \dfrac{x^{4}-x^{2}+2}{x-1}$

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

$3.\: \dfrac{x^{3}}{x+2}\\$

$Let \: \dfrac{x^{3}}{x+2} = \dfrac{x^{3}+8-8}{x+2}\\$
$=\dfrac{x^{3}+2^{3}-8}{x+2}\\$
$= \dfrac{(x+2)({x^{2}-2x+4})}{x+2}-\dfrac{8}{x+2}\\$
$= x^{2}-2x+4-\dfrac{8}{x+2}\\$
$I \: = \int \left ({x^{2}-2x+4}-\dfrac{8}{x+2} \right ) dx \\$
$\dfrac{x^{3}}{3}-\dfrac{2x^{2}}{2}+4x-8\: log \left | x+2 \right |+c \\$
$\dfrac{x^{3}}{3}-x^{2} +4x-8\: log \left | x+2 \right |+c \\$

$4. \dfrac{x^{3}+3x^{2}-7x+11}{x+5}\\$

$Let \: I = \int \: \left (\dfrac{x^{3}+3x^{2}-7x+11}{x+5} \right ) dx$

$\begin{array}{r} x^{2}-2x+3\phantom{.} \\ x+5{\overline{\smash{\big)}\,x^3+3x^{2} -7x+11\phantom{)}}}\\ \underline{-~(x^3+5x^{2})\phantom{................}}\\ -2x^{2}-7x\phantom{.........}\\ \underline{-~ (-2x^2-10x)\phantom{.......}}\\ 3x+11\phantom{.}\\ \underline{\phantom{.....}{-~ (3x+15)}}\\ -4\phantom{.}\\ \end{array}$

$\dfrac{x^{3}+3x^{2}-7x+11}{x+5} =(x^{2}-2x+3)-\dfrac{4}{x-5} \\$
$=\int \dfrac{x^{3}+3x^{2}-7x+11}{x+5} =\int\left (x^{2}-2x+3)-\dfrac{4}{x-5} \right ) dx\\$
$=\dfrac{x^{3}}{3}-\dfrac{2x^{2}}{2}+3x-4 \:log \left | x+5 \right |+c\\$
$=\dfrac{x^{3}}{3}-x^{2}+3x-4 \:log \left | x+5 \right |+c\\$

$5.\: \dfrac{3x+2}{(x-2)(x-3)}\\$

$Let \: I \: =\int \dfrac{3x+2}{(x-2)(x-3)} dx\\$
$Let \:\dfrac{3x+2}{(x-2)(x-3)}= \dfrac{A}{x-2} +\dfrac{B}{x-3}\\$
$\dfrac{3x+2}{(x-2)(x-3)}= \dfrac{A(x-3)+ B(x-2)}{(x-2)(x-3)}\\$
$3x+2 = {A(x-3)+ B(x-2)}\\$
$At \: x=3,\\$
$3(3)+2 =B(3-2)\\$
$9+2=B\\$
$B=11\\$
$At \: x=2,\\$
$3(2)+2 =A(2-3)\\$
$6+2=-A\\$
$A=-8\\$
$\dfrac{3x+2}{(x-2)(x-3)}=\dfrac{-8}{x-2}+\dfrac{11}{x-3}\\$
$=\int \dfrac{3x+2}{(x-2)(x-3)}=\dfrac{-8}{x-2}+\dfrac{11}{x-3}\\$
$=-8 \:log \left | x-2 \right |+11+\:log \left | x-2 \right |+c \\$

$6. \: \dfrac{4x^{2}+2x+6}{(x+1)^{2}(x-3)}\\$

$Let \: \dfrac{4x^{2}+2x+6}{(x+1)^{2}(x-3)}= \dfrac{A}{(x+1)}+\dfrac{B}{(x+1)^{2}}+\dfrac{C}{(x-3)} \\$
$\dfrac{4x^{2}+2x+6}{(x+1)^{2}(x-3)}= \dfrac{A(x+1)(x-3)+ B(x-3)+ C(x+1)^{2} }{(x+1)^{2}(x-3)}\\$
$4x^{2}+2x+6 = A(x+1)(x-3)+ B(x-3)+ C(x+1)^{2} \\$
$At\: x=1,\\$
$4-2+6=0+B(-1-3)+0\\$
$8=-4B\\$
$B=-2\\$
$At \:x=-3\\$
$36+6+6=0+0+C(3+1)^{2}\\$
$48=16C\\$
$C=3\\$
$At \:x=0,\\$
$0+0+6=A(0+1)(0-3)+B(0-3)+C(0-1)^{2}\\$
$6=-3A-3B+C\\$
$6=-3A-3(-2)+3\\$
$6=-3A+6+3\\$
$6-6-3=-3A\\$
$-3=-3A\\$
$A=1\\$
$\dfrac{4x^{2}+2x+6}{(x+1)^{2}(x-3)}=\dfrac{1}{(x+1)}-\dfrac{2}{(x+1)^{2}}+\dfrac{3}{(x-3)}\\$
$=\int \dfrac{4x^{2}+2x+6}{(x+1)^{2}(x-3)}=\dfrac{1}{(x+1)}-\dfrac{2}{(x+1)^{2}}+\dfrac{3}{(x-3)} dx\\$
$=\int \dfrac{1}{(x+1)}-2(x+1)^{-2}+\dfrac{3}{(x-3)} dx\\$
$=log \left | x+1 \right |+2(x+1)^{-1}+3 \: log \left | x-3 \right |+c\\$
$=log \left | x+1 \right |+\frac{2}{(x+1)}+3 \: log \left | x-3 \right |+c\\$

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

$Let \: I = \: \int \left (\dfrac{3x^{2}-2x+5}{(x-1)(x^{2}+5)} \right ) dx\\$
$\dfrac{3x^{2}-2x+5}{(x-1)(x^{2}+5)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-1)(x^{2}+5)}\\$
$3x^{2}-2x+5= A(x^{2}+5)+(Bx+C)(x-1)\\$
$At \:x=1,\\$
$3(1)^{2}-2+5=A(1+5)+0\\$
$6=6A\\$
$A=1\\$
$At \:x=0\\$
$0-0+5=A(0+5)+(0+C)(0-1)\\$
$5=5A-C\\$
$5=5(1)-C\\$
$5-5=C\\$
$C=0$
$At \: x=2\\$
$12-4+5=A(4+5)+(2B+C)(2-1)\\$
$13=9A+2B+C\\$
$13=9(1)+2B+0\\$
$13=9+2B\\$
$2B=4\\$
$B=2\\$
$\dfrac{3x^{2}-2x+5}{(x-1)(x^{2}+5)}=\dfrac{1}{(x-1)+\dfrac{2x}{(x^{2}+5)}}$
$\int \left (\dfrac{1}{x-1}+\dfrac{2x}{x^{2}+5} \right ) dx\\$
$=log \left | x-1 \right |+log \left |x^{2}+5 \right |+c\\$
$=log \left |(x-1)(x^{2}+5) \right |+c$
$=log \left |x^{3}-x^{2}+5x-5 \right |+c$

$8. If \: f'(x)\: =\dfrac{1}{x} \: and \: f(1)\: =\: \dfrac{\pi}{4},\: then f(x)\\$

$\int f'(x)\:dx=f(x)+c\\$
$\int \dfrac{1}{x} dx=f(x)+c\\$
$log \left | x \right |=f(x)+c\cdots \cdots (1)\\$
$At\: x=1\\$
$log 1 = f(1)+c\\$
$0=\dfrac{\pi}{4}+c\\$
$C=-\dfrac{\pi}{4}\\$
$In \: Eq (1) log \left | x \right |=f(x)+c\\$
$log \left | x \right |= f(x)-\dfrac{\pi}{4}\\$
$log \left | x \right |+\dfrac{\pi}{4}= f(x)\\$
$f(x)=log \left | x \right |+\dfrac{\pi}{4}\\$