Q115 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics
Given that $\int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c$.
If $\int \frac{1}{x^4+3 x^2+1} d x=a \cdot \tan ^{-1}\left(\frac{b\left(x^2-1\right)}{x}\right)$
$$
+c \tan ^{-1}\left(\frac{d\left(x^2+1\right)}{x}\right)+k
$$
where $k$ is a constant of integration, then $5(c+d+a b)=$
- Correct answerA3
- B5
- C8
- D10
Solution
we have $\int \frac{1}{x^4+3 x^2+1} d x$
now dividing by $\mathrm{x}^2$ to the numerator and denominator
$\begin{aligned}
& \Rightarrow \int \frac{\frac{1}{\mathrm{x}^2}}{\mathrm{x}^2+\frac{1}{\mathrm{x}^2}+3} \mathrm{dx} \\
& \Rightarrow \int \frac{\frac{1}{2}\left\{\left(1+\frac{1}{\mathrm{x}^2}\right)-\left(1-\frac{1}{\mathrm{x}^2}\right)\right\} \mathrm{dx}}{\mathrm{x}^2+\frac{1}{\mathrm{x}^2}+3} \\
& \Rightarrow \frac{1}{2} \int \frac{1+\frac{1}{\mathrm{x}^2}}{\left(\mathrm{x}^2+\frac{1}{\mathrm{x}^2}+3\right)} \mathrm{dx}-\frac{1}{2} \int \frac{1-\frac{1}{\mathrm{x}^2}}{\mathrm{x}^2+\frac{1}{\mathrm{x}^2}+3} \mathrm{dx} \\
& \Rightarrow \frac{1}{2} \int \frac{1}{\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^2+5} \mathrm{dx}-\frac{1}{2} \int \frac{1}{\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^2+1} \mathrm{dx} \\
& \Rightarrow \frac{1}{2} \cdot \frac{1}{\sqrt{5}} \tan ^{-1}\left(\frac{\left.\mathrm{x}-\frac{1}{\mathrm{x}}\right)}{\sqrt{5}}\right)-\frac{1}{2} \tan ^{-1}\left(\frac{\mathrm{x}+\frac{1}{\mathrm{x}}}{1}\right)+\mathrm{c}
\end{aligned}$
On comparing with R.H.S
$\operatorname{atan}^{-1}\left(\frac{\mathrm{b}\left(\mathrm{x}^2-1\right)}{\mathrm{x}}\right)+\mathrm{C} \tan ^{-1}\left(\frac{\mathrm{d}\left(\mathrm{x}^2+1\right)}{\mathrm{x}}\right)+\mathrm{k}$
$\Rightarrow$ we get $a=\frac{1}{2 \sqrt{5}}, \mathrm{~b}=\frac{1}{\sqrt{5}}, \mathrm{c}=\frac{-1}{2}, \mathrm{~d}=1$
$\Rightarrow 5(\mathrm{c}+\mathrm{d}+\mathrm{ab})=3$
Hence, the correct answer is option: 'A'.