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Prove the following
$\int_1^3 \frac{d x}{x^2(x+1)}=\frac{2}{3}+\log \frac{2}{3}$
$\int_1^3 \frac{d x}{x^2(x+1)}=\frac{2}{3}+\log \frac{2}{3}$
Solution:
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Verified Answer
Let $\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}$ $\therefore \quad 1 \equiv \mathrm{Ax}(\mathrm{x}+1)+\mathrm{B}(\mathrm{x}+1)+\mathrm{Cx}^2$
Put $x=0, \Rightarrow B=1 \quad$ Put $x=-1 \Rightarrow C=1$
Comparing the coefficient $\Rightarrow \mathrm{A}=-\mathrm{C}=-1$
$\begin{aligned}
&\text { LHS }=-\int_1^3 \frac{1}{x} d x+\int_1^3 \frac{1}{x^2} d x+\int_1^3 \frac{1}{x+1} d x \\
&=-[\log x]_1^3+\left[-\frac{1}{x}\right]_1^3+[\log (x+1)]_1^3 \\
&=\frac{2}{3}+\log \frac{2}{3}=\text { R.H.S. }
\end{aligned}$
Put $x=0, \Rightarrow B=1 \quad$ Put $x=-1 \Rightarrow C=1$
Comparing the coefficient $\Rightarrow \mathrm{A}=-\mathrm{C}=-1$
$\begin{aligned}
&\text { LHS }=-\int_1^3 \frac{1}{x} d x+\int_1^3 \frac{1}{x^2} d x+\int_1^3 \frac{1}{x+1} d x \\
&=-[\log x]_1^3+\left[-\frac{1}{x}\right]_1^3+[\log (x+1)]_1^3 \\
&=\frac{2}{3}+\log \frac{2}{3}=\text { R.H.S. }
\end{aligned}$
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