Evalute : 3 x-2 (x+1)^2(x+3) d x - Vidyadip

Question

Evalute : $\int \frac{3 x-2}{(x+1)^2(x+3)} d x$

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Answer

Let $I =\int \frac{3 x-2}{(x+1)^2(x+3)} d x$
Let $\frac{3 x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3}$
$
\therefore 3 x -2= A ( x +1)( x +3)+ B ( x +3)+ C ( x +1)^2
$
Put $x+1=0$, i.e. $x=-1$, we get
$
3(-1)-2= A (0)(2)+ B (2)+ C (0)
$
$
\therefore-5=2 B
$
$
\therefore B =-\frac{5}{2}
$
Put $x+3=0$, i.e. $x=-3$, we get
$
\begin{aligned}
& 3(-3)-2= A (-2)(0)+ B (0)+ C (4) \\
& \therefore-11=4 C \\
& \therefore C =-\frac{11}{4}
\end{aligned}
$
Put $x =0$, we get
$
\begin{aligned}
& 3(0)-2= A (1)(3)+ B (3)+ C (1) \\
& \therefore-2=3 A +3 B + C
\end{aligned}
$
But $B =-\frac{5}{2}$ and $C =-\frac{11}{4}$
$
\begin{aligned}
& \therefore-2=3 A+3\left(-\frac{5}{2}\right)-\frac{11}{4} \\
& \therefore 3 A=-2+\frac{15}{2}+\frac{11}{4}=\frac{-8+30+11}{4}=\frac{33}{4} \\
& \therefore A=\frac{11}{4}\\
& \therefore \frac{3 x-2}{(x+1)^2(x+3)}=\frac{\left(\frac{11}{4}\right)}{x+1}+\frac{\left(-\frac{5}{2}\right)}{(x+1)^2}+\frac{\left(-\frac{11}{4}\right)}{(x+3)}
\end{aligned}
$
$\begin{aligned} \therefore I & =\int\left[\frac{\left(\frac{11}{4}\right)}{x+1}+\frac{\left(-\frac{5}{2}\right)}{(x+1)^2}+\frac{\left(-\frac{11}{4}\right)}{x+3}\right] d x \\ & =\frac{11}{4} \int \frac{1}{x+1} d x-\frac{5}{2} \int(x+1)^{-2} d x-\frac{11}{4} \int \frac{1}{x+3} d x \\ & =\frac{11}{4} \log |x+1|-\frac{5}{2} \cdot \frac{(x+1)^{-1}}{-1}-\frac{11}{4} \log |x+3|+c \\ & =\frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+c .\end{aligned}$

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