Q117 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Mathematics
$\int \frac{d x}{\sqrt{\left(5+2 x+x^2\right)^3}}$ is equal to
- A$\frac{1}{4} \frac{1}{\sqrt{5+2 x+x^2}}+C$
- B$\frac{1}{\sqrt{5+2 x+x^2}}+C$
- C$\frac{x+1}{\sqrt{5+2 x+x^2}}+C$
- Correct answerD$\frac{1}{4} \frac{x+1}{\sqrt{5+2 x+x^2}}+C$
Solution
$5+2 x+x^2=4+(x+1)^2$
Let $x+1=z \Rightarrow d x=d z$
$$
\therefore \quad I=\int \frac{1 \cdot d x}{\sqrt{\left(5+2 x+x^2\right)^3}}=\int \frac{d z}{\sqrt{\left(4+z^2\right)^3}}
$$
Let $z=2 \tan t, d z=\frac{2 d t}{\cos ^2 t}$
$$
\sqrt{4+z^2}=2 \sqrt{1+\tan ^2 t}=\frac{2}{\cos t}
$$
So, $I=\frac{1}{4} \int \cos t d t=\frac{1}{4} \sin t+C$
$$
=\frac{1}{4} \frac{\tan t}{\sqrt{1+\tan ^2 t}}+C=\frac{1}{4} \frac{z^2}{\sqrt{1+\frac{z^2}{4}}}+C
$$
$$
\Rightarrow I=\frac{x+1}{4 \sqrt{5+2 x+x^2}}+C
$$
Hence, the correct answer is option: 'D'.