Question

Prove the following question.

${\int }_1^3\frac{1}{x^2(x+1)}dx=\frac{2}{3}+log\frac{2}{3}$

Answer 1

${\int }_1^3\frac{1}{x^2(x+1)}dx=\frac{2}{3}+log\frac{2}{3}$

By using partial fraction method

$Let\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}..\left(i\right)\Rightarrow 1=Ax(x+1)+B(x+1)+C{x}^2..\left(ii\right)$

Substituting x = 0, -1 in Eq. (ii), we get

1 = B(0+1) and 1 = C$(-1{)}^2\Rightarrow$ B = 1 and C = 1

On equating the coefficients of $x^2$ on both sides of Eq.(ii), we get

0 = A + C $\Rightarrow$ A = -C = -1

$\therefore {\int }_1^3\frac{1}{x^2(x+1)}dx={\int }_1^3(-\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x+1})dx={[-log|x|-\frac{1}{x}+log|x+1|]}_1^3={[log\left|\frac{x+1}{x}\right|-\frac{1}{x}]}_1^3=\left(log|\frac{4}{3}-\frac{1}{3}|\right)-(log\left|\frac{2}{1}\right|-\frac{1}{1})=(log\frac{4}{3}-log2)+(1-\frac{1}{3})[\because log\left(\frac{m}{n}\right)=logm-logn]=log(\frac{4}{3}\times \frac{1}{2})+\frac{2}{3}=log\frac{2}{3}+\frac{2}{3}Henceproved.$