Answer to Question #104047 in Calculus for Patrick Kariuki

Question

Integrate (7x+8)/(2x²+11x+5)dx
Integrate (3x²+18x+3)/(3x²+5x-2)dx
Integrate (6x²)/(x²-4)dx
Integrate (sin inverse x) /(x²-4)dx
Integrate (x²+1)³/(x+2)dx

Answer 1

1. $I=\int\frac{7x+8}{2x^2+11x+5}dx$

decompose the denominator into factors

$2x^2+11x+5=2(x+5)(x+0.5)\\ 2x^2+11x+5=0\\ D=121-4\cdot2\cdot5=81\\ x_1=\frac{-11-9}{4}=-5\\ x_2=\frac{-11+9}{4}=-0.5$

$I=\int\frac{7x+8}{2(x+5)(x+0.5)}dx=\\ =\frac{1}{2}\int\frac{7x+8}{(x+5)(x+0.5)}dx$

decompose into simple fractions

$\frac{7x+8}{(x+5)(x+0.5)}=\frac{A}{(x+5)}+\frac{B}{(x+0.5)}=\frac{A(x+0.5)+B(x+5)}{(x+5)(x+0.5)}\\ 1) x=-5: \\ 7\cdot(-5)+8=A(-5+0.5)\\ A=6\\ 2) x=-0.5\\ 7\cdot(-0.5)+8=B(-0.5+5)\\ B=1$

$I=\frac{1}{2}\int(\frac{6}{x+5}+\frac{1}{x+0.5})dx=\\ =\frac{1}{2}(6ln|x+5|+ln|x+0.5|)+C$

2

$I=\int\frac{3x^2+18x+3}{3x^2+5x-2}dx$

perform the division and decompose the denominator into factors

$\begin{matrix} 3x^2+18x+3| & 3x^2+5x-2 \\ 3x^2+5x-2 & 1\\ -----\\ 13x+5\\ \end{matrix}\\ \frac{3x^2+18x+3 }{3x^2+5x-2} =1+\frac{13x+5}{3x^2+5x-2 }\\ 3x^2+5x-2 =3(x+2)(x-\frac{1}{3})\\ 3x^2+5x-2 =0\\ D=25+4\cdot3\cdot2=49\\ x_1=\frac{-5-7}{6}=-2\\ x_2=\frac{-5+7}{6}=\frac{1}{3}$

$I=\int(1+\frac{13x+5}{3(x+2)(x-\frac{1}{3})})dx=\\ =x+\frac{1}{3}\int\frac{13x+5}{(x+2)(x-\frac{1}{3})}$

decompose into simple fractions

$\frac{13x+5}{(x+2)(x-\frac{1}{3})}=\frac{A}{(x+2)}+\frac{B}{(x-\frac{1}{3})}=\frac{A(x-\frac{1}{3})+B(x+2)}{(x+2)(x-\frac{1}{3})}\\ 1) x=-2: \\ 13\cdot(-2)+5=A(-2-\frac{1}{3})\\ A=\frac{33}{7}\\ 2) x=\frac{1}{3}\\ 13\cdot\frac{1}{3}+5=B(\frac{1}{3}+2)\\ B=4$

$I=x+\frac{1}{3}\int(\frac{\frac{33}{7}}{x+2}+\frac{4}{x-\frac{1}{3}})dx=\\ =x+\frac{1}{3}(\frac{33}{7}ln|x+2|+4ln|x-\frac{1}{3}|)+C$

3

$\int\frac{\sin^{-1}x}{x^2-4}dx=\frac{1}{8}(-2i(Li_2(i(-2+\sqrt{3})e^{-i\sin^{-1}x})+\\ +Li_2(-i(2+\sqrt{3})e^{-i\sin^{-1}x}))+\\ +2i(Li_2(-i(-2+\sqrt{3})e^{-i\sin^{-1}x})+\\ +Li_2(-i(2+\sqrt{3})e^{-i\sin^{-1}x}))+\\ +(-2\sin^{-1}x+\pi+4sin^{-1}\sqrt{\frac{3}{2}})\cdot\\ \cdot \log(1-i(\sqrt{3}-2)e^{-i\sin^{-1}x})+\\ +(-2\sin^{-1}x+\pi+4sin^{-1}\sqrt{\frac{3}{2}})\cdot\\ \cdot \log(1+i(\sqrt{3}+2)e^{-i\sin^{-1}x})+\\ +\log(x-2)(\pi-2\sin^{-1}x)+\\ +2\log(x-2)\sin^{-1}x-\\ -\log(x+2)(\pi-2\sin^{-1}x)-\\ -2\log(x+2)\sin^{-1}x-\\ -\log(1+i(\sqrt{3}-2)e^{-i\sin^{-1}x})\cdot\\ \cdot(-2\sin^{-1}x+\pi-4i\sinh^{-1}\frac{1}{\sqrt{2}})-\\ --\log(1-i(\sqrt{3}+2)e^{-i\sin^{-1}x})\cdot\\ \cdot(-2\sin^{-1}x+\pi+4i\sinh^{-1}\frac{1}{\sqrt{2}})+\\ +8isin^{-1}\sqrt{\frac{3}{2}} \tan^{-1}(\frac{\cot(\frac{1}{4}(2\sin^{-1}x+\pi))}{\sqrt{3}})-\\ - 8\sinh^{-1}\frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{3}\cot(\frac{1}{4}(2\sin^{-1}x+\pi))))+c$

$\sin^{-1}x$ is the inverse sine function

$Li_nx$ isthe polylogarithm function

4

$\int\frac{6-x}{x^2-4}dx=\int\frac{6}{x^2-4}dx-\int\frac{x}{x^2-4}dx=\\ =6\cdot\frac{1}{4}\ln|\frac{x-2}{x+2}| -\frac{1}{2}\int\frac{d(x^2-4)}{x^2-4}dx=\\ =\frac{3}{2}\ln|\frac{x-2}{x+2}| -\frac{1}{2}\ln|x^2-4|+C$

5

$I=\int\frac{(x^2+1)^3}{x+2}dx=\int\frac{x^6+3x^4+3x^2+1}{x+2}dx$

decompose the numerator by degrees $x+2$

$\begin{matrix} & 1&0&3&0&3&0&1 \\ -2 & 1&-2&7&-14&31&-62&125\\ -2&1&-4&15&-44&119&-300\\ -2&1&-6&27&-98&315\\ -2&1&-8&43&-184\\ -2&1&-10&63\\ -2&1&-12\\ -2&1 \end{matrix}$

then

$(x+2)^6-12(x+2)^5+63(x+2)^4-\\ -184(x+2)^3+315(x+2)^2-300(x+2)+125$

$I=\int\frac{(x+2)^6-12(x+2)^5+63(x+2)^4-184(x+2)^3+315(x+2)^2-300(x+2)+125}{x+2}dx=\\ =\int((x+2)^5-12(x+2)^4+63(x+2)^3-\\ -184(x+2)^2+315(x+2)-300+\frac{125}{x+2})dx=\\ =\frac{(x+2)^6}{6}-12\cdot\frac{(x+2)^5}{5}+63\cdot\frac{(x+2)^4}{4}-\\ -184\cdot\frac{(x+2)^3}{3}+315\cdot\frac{(x+2)^2}{2}-300x+\\ +125ln|x+2|+C$