Q105 / 180 · KCET 2022 · Mathematics
If $\int \frac{d x}{(x+2)\left(x^2+1\right)}=a \log \left|1+x^2\right|+b \tan ^{-1} x$ $+\frac{1}{5} \log |x+2|+c$, then
- Correct answerA$a=\frac{-1}{10}, b=\frac{2}{5}$
- B$a=\frac{1}{10}, b=\frac{2}{5}$
- C$a=\frac{-1}{10}, b=\frac{-2}{5}$
- D$a=\frac{1}{10}, b=\frac{-2}{5}$
Solution
Given,
$$
\begin{aligned}
& \int \frac{d x}{(x+2)\left(x^2+1\right)}=a \log \left|1+x^2\right|+b \tan ^{-1} x \\
& +\frac{1}{5} \log |x+2|+c \\
& \text { Let } \frac{1}{(x+2)\left(x^2+1\right)}=\frac{A}{(x+2)}+\frac{B x+C}{\left(x^2+1\right)} \\
& \therefore \quad 1=A\left(x^2+1\right)+(x+2)(B x+C) \\
& 1=A x^2+A+B x^2+C x+2 B x+2 C \\
& 1=(A+B) x^2+x(2 B+C)+A+2 C \\
&
\end{aligned}
$$
Here, $A+B=0$, ;
$$
2 B+C=0 \text { and } A+2 C=1
$$
On solving them, we get
$$
A=\frac{1}{5}, B=\frac{-1}{5} \text { and } C=\frac{2}{5}
$$
$$
\begin{aligned}
& \text { Therefore, } \int \frac{d x}{(x+2)\left(x^2+1\right)} \\
& =\int\left(\frac{1}{5(x+2)}+\frac{-\frac{5}{5} x+\frac{2}{5}}{x^2+1}\right) d x \\
& =\frac{1}{5} \int \frac{1}{(x+2)} d x-\frac{1}{5} \int \frac{x}{x^2+1} d x+\frac{2}{5} \int \frac{d x}{1+x^2} \\
& =\frac{1}{5} \log |x+2|-\frac{1}{10} \log \left|1+x^2\right|+\frac{2}{5} \tan ^{-1} x+c \\
& =-\frac{1}{10} \log \left(1+x^2\right)+\frac{2}{5} \tan ^{-1} x+\frac{1}{5} \log |x+2|+c \ldots
\end{aligned}
$$
On comparing Eqs. (i) and (ii), we get
$$
a=\frac{-1}{10} \text { and } b=\frac{2}{5}
$$
Hence, the correct answer is option: 'A'.