# Evaluate the following. 1. \int \frac{10}{(x - 2)^2 (x^2 + 2x + 2)} dx 2. \int^\infty_1...

## Question:

Evaluate the following.

1. {eq}\int \frac{10}{(x - 2)^2 (x^2 + 2x + 2)} dx {/eq}

2. {eq}\int^\infty_1 \frac{1}{x(x + 1)} dx {/eq}

## Integration:

The above question concerns the topic of the integration by partial fraction. following is the partial fraction form for integration

{eq}\displaystyle \frac{1}{(x+a)(x^2+bx+c)}=\frac{A}{x+a}+\frac{Bx+C}{x^2+bx+c}\\ {/eq}

By taking LCM on both sides and comparing the coefficient, we get the value of A, B, C.

After doing partial fraction, given integral is converted into simple form, which can be evaluated directly.

1.

{eq}\begin{align} \displaystyle &\int \frac{10}{(x - 2)^2 (x^2 + 2x + 2)} dx\\ &\text{First we do partial fraction of given function so }\\ \displaystyle \frac{10}{(x - 2)^2 (x^2 + 2x + 2)} &= \frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{Cx+D}{x^2+2x+2}\\ \displaystyle \frac{10}{(x - 2)^2 (x^2 + 2x + 2)} &=\frac{A(x-2)(x^2+2x+2)+B(x^2+2x+2)+(Cx+D)(x-2)^2}{(x-2)^2 (x^2+2x+2)}\\ \displaystyle 10 &=A(x^3-2x-4)+B(x^2+2x+2)+(Cx+D)(x^2-4x+4)\\ \displaystyle 10 &=A(x^3-2x-4)+B(x^2+2x+2)+Cx^3-4Cx^2+4Cx+Dx^2-4Dx+4D\\ \displaystyle 10 &=x^3(A+C)+x^2(B-4C+D)+x(-2A+2B+4C-4D)+(-4A+2B+4D)\\ &\text{Comparing both side, we have}\\ A+C&=0 ---(1)\\ B-4C+D &=0---(2)\\ -2A+2B+4C-4D &=0---(3)\\ -4A+2B+4D &=10---(4)\\ \text{from equation (1), we have } A &=-C \text{ and from equation(2), we have }B=4C-D\\ &\text{Substituting the value of A and B in equation(3) and (4), we have }\\ -2(-C)+2(4C-D)+4C-4D=0 &\rightarrow 14C-6D =0 ---(5)\\ -4(-C)+2(4C-D)+4D=10 &\rightarrow 12C+2D =10---(6)\\ \text{Solving equation(5) and (6), we have}\\ C=\frac{3}{5}, D=\frac{7}{5}\\ A=-C= \frac{3}{5}, B=4C-D=4(\frac{3}{5})-\frac{7}{5}=1\\ \end{align} {/eq}

So the given integral is converted into

{eq}\begin{align} \displaystyle \int \frac{10}{(x - 2)^2 (x^2 + 2x + 2)} dx &=\int \frac{-3}{5(x-2)}+\frac{1}{(x-2)^2}+\frac{3x+7}{5(x^2+2x+2)} dx \\ \text{Let } I &= I_1+I_2+I_3 ---(7)\\ \displaystyle \text{Where }I_1 &=\int \frac{-3}{5(x-2)}dx= \frac{-3}{5} \log (x-2)---(8)\\ \displaystyle \text{and }I_2 &= \int \frac{1}{(x-2)^2} dx= \frac{(x-2)^{-2+1}}{-2+1}\\ \displaystyle I_2 &=-\frac{1}{x-2} ---(9)\\ \displaystyle \text{and }I_3 &= \int \frac{3x+7}{5(x^2+2x+2)} dx \\ \displaystyle I_3 &=\int\frac{3x+7}{5(x^2+2x+2)}dx \\ \displaystyle I_3 &=\frac{3}{5}\int \frac{x+\frac{7}{3}}{(x^2+2x+2)}dx\\ \displaystyle I_3 &=\frac{3}{10}\int \frac{2x+\frac{14}{3}+2-2}{(x^2+2x+2)}dx\\ \displaystyle I_3 &=\frac{3}{10}\left [ \int \frac{2x+2}{x^2+2x+2} dx+ \int \frac{8}{3(x^2+2x+2)}dx\right ]\\ \displaystyle I_3 &=\frac{3}{10}\left [ I_4+I_5 \right ]\\ \displaystyle \text{Now }I_4 &=\int \frac{2x+2}{x^2+2x+2} dx\\ \displaystyle \text{Let } x^2+2x+2 &=t \rightarrow (2x+2)dx=dt\\ \displaystyle I_4 &=\int \frac{dt}{t}=\log t+C\\ \displaystyle I_4 &=\log (x^2+2x+2)---(10)\\ \displaystyle \text{Now }I_5 &= \int \frac{8}{3(x^2+2x+2)}dx\\ \displaystyle I_5 &= \int \frac{8}{3(x^2+2x+4+2-4)}dx\\ \displaystyle I_5 &= \int \frac{8}{3[(x+2)^2-(\sqrt{2})^2)]}dx\\ \displaystyle I_5 &=\frac{8}{3} \left [ \frac{1}{2\sqrt(2)}\log \left (\frac{x-\sqrt(2)}{x+\sqrt(2)} \right ) \right ]---(11) &\left [ \int \frac{1}{x^2-a^2} =\frac{1}{2a}\log \left ( \frac{x-a}{x+a} \right )\right ]\\ \displaystyle \text{Now }I_3 &=\frac{3}{10}\left [\log (x^2+2x+2)+\frac{8}{3} \left [ \frac{1}{2\sqrt(2)}\log \left (\frac{x-\sqrt(2)}{x+\sqrt(2)} \right ) \right ]- \right ]---(12)\\ &\text{Substituting the value of }I_1,I_2,I_3 \text{ in equation (7), we have}\\ \displaystyle I &=\int \frac{-3}{5(x-2)}dx= \frac{-3}{5} \log (x-2)-\frac{1}{x-2}+\frac{3}{10}\left [\log (x^2+2x+2)+\frac{8}{3} \left [ \frac{1}{2\sqrt(2)}\log \left (\frac{x-\sqrt(2)}{x+\sqrt(2)} \right ) \right ] \right ]\\ \end{align} {/eq}

2.

{eq}\begin{align} \displaystyle &\int^\infty_1 \frac{1}{x(x + 1)} dx\\ &\text{First we do partial fraction of given function so }\\ \displaystyle \frac{1}{x(x + 1)} &=\frac{A}{x}+\frac{B}{x+1}\\ \displaystyle \frac{1}{x(x + 1)} &=\frac{Ax+A+Bx}{x(x+1)}\\ \displaystyle 1 &=(A+B)x+A\\ &\text{Comparing both side, we have}\\ A=1, A+B=0 &\rightarrow B=-A=-1 \text{ so}\\ \displaystyle \int^\infty_1 \frac{1}{x(x + 1)} dx &= \int^\infty_1 \left (\frac{1}{x} -\frac{1}{x+1} \right )dx\\ \displaystyle &=\left ( \log (x)- \log (x+1) \right )^\infty_1\\ \displaystyle &= \left [\log \left ( \frac{x}{x+1}\right ) \right ]^\infty_1 & \left [ \log a-\log b= \log \frac{a}{b} \right ]\\ \displaystyle &=\left [\log \left ( \frac{1}{1+\frac{1}{x}}\right ) \right ]^\infty_1\\ \displaystyle &=\left [\log \left ( \frac{1}{1+\frac{1}{\infty}}\right ) -\log \left ( \frac{1}{1+\frac{1}{1}}\right ) \right ]\\ \displaystyle &=\left [ \log 1- \log \frac{1}{2} \right ]\\ \displaystyle &=\log 2 &\left [ -\log\frac{1}{a} =\log a\right ] \end{align} {/eq}