Evalute : 5 x^2+20 x+6 x^3+2 x^2+x d x

Question

Evalute : $\int \frac{5 x^2+20 x+6}{x^3+2 x^2+x} d x$

✓

Answer

Let $I=\int \frac{5 x^2+20 x+6}{x^3+2 x^2+x} d x$
$
\begin{aligned}
& =\int \frac{5 x^2+20 x+6}{x\left(x^2+2 x+1\right)} d x \\
& =\int \frac{5 x^2+20 x+6}{x(x+1)^2} d x
\end{aligned}
$
Let $\frac{5 x^2+20 x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$
$
\therefore 5 x ^2+20 x +6= A ( x +1)^2+ Bx ( x +1)+ Cx
$
Put $x=0$, we get
$
\begin{aligned}
& 0+0+6= A (1)+ B (0)(1)+ C (0) \\
& \therefore A =6
\end{aligned}
$
Put $x+1=0$, i.e. $x=-1$, we get
$
\begin{aligned}
& 5(1)+20(-1)+6= A (0)+ B (-1)(0)+ C (-1) \\
& \therefore-9=- C \\
& \therefore C =9
\end{aligned}
$
Put $x =1$, we get
$
\begin{aligned}
& 5(1)+20(1)+6=A(4)+B(1)(2)+C(1) \\
& \text { But } A=6 \text { and } C=9 \\
& \therefore 31=24+2 B+9 \\
& \therefore B=-1 \\
& \therefore \frac{5 x^2+20 x+6}{x(x+1)^2}=\frac{6}{x}-\frac{1}{x+1}+\frac{9}{(x+1)^2} \\
& \therefore I=\int\left[\frac{6}{x}-\frac{1}{x+1}+\frac{9}{(x+1)^2}\right] d x
\end{aligned}
$
$
\begin{aligned}
& =6 \int \frac{1}{x} d x-\int \frac{1}{x+1} d x+9 \int(x+1)^{-2} d x \\
& =6 \log |x|-\log |x+1|+9 \cdot \frac{(x+1)^{-1}}{-1}+c
\end{aligned}
$
$
=6 \log |x|-\log |x+1|-\frac{9}{x+1}+c .
$

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