Q59 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics
$\frac{d}{d x}\left(\frac{x+5}{(x+1)^2(x+2)}\right)=$
- A$\frac{8}{(x+2)^2}-\frac{3}{(x+1)^2}+\frac{3}{(x+1)^3}$
- Correct answerB$\frac{3}{(x+1)^2}-\frac{3}{(x+2)^2}-\frac{8}{(x+1)^3}$
- C$\frac{3}{(x+2)^2}-\frac{3}{(x+1)^3}-\frac{8}{(x+1)^2}$
- D$\frac{8}{(x+2)^2}-\frac{3}{(x+1)^3}+\frac{3}{(x+1)^2}$
Solution
Let $\frac{x+5}{(x+1)^2(x+4)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{(x+4)}$
$$
\begin{aligned}
& \Rightarrow x+5=A(x+1)(x+2)+B(x+2)+c(x+1)^2 \\
& \Rightarrow x+5=x^2(A+C)+x(3 A+B+2 C)+2 A+2 B+C .
\end{aligned}
$$
On comparing the coefficients of like power of $x$, we get
$$
\begin{aligned}
A+C & =0, \\
3 A+B+2 C & =1 \text { and } 2 A+2 B+C=5
\end{aligned}
$$
On substituting $A=-C$ in last two equations, we get
$$
\begin{aligned}
B-C & =1 \\
2 B-C & =5
\end{aligned}
$$
and
$$
2 B-C=5
$$
On subtracting Eqs. (ii) and (iii), we get
$$
B=4 \Rightarrow C=3
$$
[from Eq. (ii)]
and hence $A=-3$
Now, $\frac{(x+5)}{(x+1)^2(x+2)}=\frac{-3}{(x+1)}+\frac{4}{(x+1)^2}+\frac{3}{(x+2)}$
On differentiating both sides w.r.t $x$, we get.
$$
\frac{d}{d x}\left(\frac{(x+5)}{(x+1)^2(x+4)}\right)=\frac{3}{(x+1)^2}-\frac{8}{(x+1)^3}-\frac{3}{(x+2)^2}
$$
Hence, the correct answer is option: 'B'.