Q109 / 160 · AP EAMCET 2023 (16 May Shift 1) · Mathematics
$\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x=$
- A$\frac{1}{2 \mathrm{x}^2} \sqrt{2 \mathrm{x}^4+2 \mathrm{x}^2+1}+\mathrm{C}$
- Correct answerB$\frac{1}{2 x^2} \sqrt{2 x^4-2 x^2+1}+C$
- C$\frac{1}{2 \mathrm{x}^2} \sqrt{4 \mathrm{x}^4-2 \mathrm{x}^2+1}+\mathrm{C}$
- D$\frac{1}{2 x^2} \sqrt{4 x^4+2 x^2+1}+C$
Solution
$\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x=\int \frac{\frac{x^2-1}{x^5}}{\frac{x^3 \sqrt{2 x^4-2 x^2+1}}{x^5}} d x$
$=\int \frac{\frac{1}{x^3}-\frac{1}{x^5}}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}} d x$ ...(i)
Let $2-\frac{2}{x^2}+\frac{1}{x^4}=t \Rightarrow\left(0+\frac{4}{x^3}-\frac{4}{x^5}\right) d x=d t$
$\Rightarrow\left(\frac{1}{x^3}-\frac{1}{x^5}\right) d x=\frac{d t}{4}$
Substituting above values in equation (i), we get
$\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x=\int \frac{d t}{4 \sqrt{t}}$
$=\frac{1}{4} \cdot \frac{t^{1 / 2}}{\left(\frac{1}{2}\right)}+C=\frac{1}{2}\left(2-\frac{2}{x^2}+\frac{1}{x^4}\right)^{1 / 2}+C$
$=\frac{1}{2 x^2} \sqrt{2 x^4-2 x^2+1}+C$.
Hence, the correct answer is option: 'B'.