Q114 / 160 · AP EAMCET 2018 (22 Apr Shift 1) · Mathematics
$\int \frac{d x}{x^3+3 x^2+2 x}=$
- A$\log |x|+\log \left|\frac{x+2}{x+1}\right|+c$
- B$\log |x|-\log |x+1|+\log |x+2|+c$
- C$\frac{1}{2}[\log |x|+\log |x+1|+\log |x+2|]+c$
- Correct answerD$\frac{1}{2} \log \left(\frac{\left|x^2+2 x\right|}{(x+1)^2}\right)+c$
Solution
Let $\quad \begin{aligned} I & =\int \frac{d x}{x^3+3 x^2+2 x} \\ & =\int \frac{d x}{x\left(x^2+3 x+2\right)} \\ & =\int \frac{d x}{x(x+1)(x+2)}\end{aligned}$
$$
\begin{aligned}
& \text { Let } \frac{1}{x(x+1)(x+2)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2} \\
& \Rightarrow 1=A(x+1)(x+2)+B x(x+2)+C x(x+1)
\end{aligned}
$$
Put $x=0$, we get
$$
A=\frac{1}{2}
$$
Put $x=-1$, we get
$$
B=-1
$$
Put $x=-2$, we get
$$
\begin{aligned}
& \quad C=\frac{1}{2} \\
& \therefore \quad I=\int\left(\frac{1}{2 x}-\frac{1}{x+1}+\frac{1}{2(x+2)}\right) d x \\
& =\frac{1}{2} \log x-\log (x+1)+\frac{1}{2} \log (x+2)+C \\
& =\frac{1}{2}[\log x-2 \log (x+1)+\log (x+2)]+C \\
& =\frac{1}{2} \log \left|\frac{x(x+2)}{(x+1)^2}\right|+C \\
& =\frac{1}{2} \log \left|\frac{x^2+2 x}{(x+1)^2}\right|+C .
\end{aligned}
$$
Hence, the correct answer is option: 'D'.