Question

Evaluate : $\int \frac{3x+1}{{\left(x-2\right)}^2\left(x+2\right)}dx$

• A. $\frac{5}{16}log\left|x-2\right|+\frac{7}{4\left(x-2\right)}+\frac{5}{16}log\left|x+2\right|+C$
• B. $\frac{5}{16}log\left|\left(x-2\right)\right|+\frac{3}{4\left(x-2\right)}+\frac{5}{16}log\left|x-2\right|+C$
• C. $\frac{5}{16}log\left|x-2\right|-\frac{7}{4\left(x-2\right)}-\frac{5}{16}log\left|x+2\right|+C$
• D. None of these

Answer 1

Answer: (C)

Let $\frac{3x+1}{{\left(x-2\right)}^2\left(x+2\right)}=\frac{A}{x-2}+\frac{B}{{\left(x-2\right)}^2}+\frac{C}{x+2}............\left(i\right)$

$\Rightarrow 3x+1=A\left(x-2\right)\left(x+2\right)+B\left(x+2\right)+C{\left(x-2\right)}^2.....\left(ii\right)$

Put $x=2$ in (ii), we get $7=4B\Rightarrow B=\frac{7}{4}$

Put $x=-2$ in (ii), we get $-5=16C\Rightarrow C=-\frac{5}{16}$

Comparing coefficients of $x^2$ on both sides of $\left(ii\right)$, we get $A+C=0\Rightarrow A=-C\Rightarrow A=\frac{5}{16}$

$\therefore I=\int \frac{3x+1}{{\left(x-2\right)}^2\left(x+2\right)}dx$

$=\frac{5}{16}\int \frac{1}{x-2}dx+\frac{7}{4}\int \frac{1}{{\left(x-2\right)}^2}dx-\frac{5}{16}\int \frac{1}{x+2}dx$

$\Rightarrow I=\frac{5}{16}log\left|x-2\right|-\frac{7}{4\left(x-2\right)}-\frac{5}{16}log\left|x+2\right|+C$