Question

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

Answer 1

By the method of partial fractions,

$\frac{3x+1}{{(x+1)}^2(x+3)}=\frac{A}{x+1}+\frac{B}{{(x+1)}^2}+\frac{C}{x+3}$

$\Rightarrow 3x+1=A(x+1)(x+3)+B(x+3)+C{(x+1)}^2$

Put $x=-1$

$\Rightarrow 3\times -1+1=B(-1+3)$

$\Rightarrow -2=2B$ or $B=-1$

Put $x=-3$

$\Rightarrow 3\times -3+1=C{(-3+1)}^2$

$\Rightarrow -9+1=4C$

$\Rightarrow 4C=-8$ or $C=-2$

Put $x=0$

$\Rightarrow 3\times 0+1=A(0+1)(0+3)+B(0+3)+C{(0+1)}^2$

$\Rightarrow 3A+3B+C=1$

$\Rightarrow 3A+3\times -1-2=1$

$\Rightarrow 3A=1+5=6$

$\therefore A=\frac{6}{3}=2$

Now $\frac{3x+1}{{(x+1)}^2(x+3)}=\frac{2}{x+1}-\frac{1}{{(x+1)}^2}-\frac{2}{x+3}$

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

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

$=2\log |x+1|-\frac{{(x+1)}^{-2+1}}{-2+1}-2\log |x+3|+c$

$=2\log \left|\frac{x+1}{x+3}\right|+\frac{1}{x+1}+c$ where $c$ is the constant of integration.

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