Question

$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=$

• A. $-\frac{1}{2}\sqrt{2–\frac{2}{x^2}+\frac{1}{x^4}}+C$
• B. $\frac{1}{2}\log (2x^4–2x^2+1)+C$
• C. $\frac{1}{2}\sqrt{2–\frac{2}{x^2}+\frac{1}{x^4}}+C$
• D. None of these

Answer 1

The correct option is C

$\frac{1}{2}\sqrt{2–\frac{2}{x^2}+\frac{1}{x^4}}+C$

Explanation for the correct option:

Evaluating the integration:

$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx$

$$\begin{gathered} \Rightarrow \int \frac{x^2-1}{x^3\sqrt{x^4\left(2-{\displaystyle \frac{2}{x^2}}+{\displaystyle \frac{1}{x^4}}\right)}}dx[\text{factor out}{x}^4\text{in denominator}] \\ \Rightarrow \int \frac{x^2-1}{x^5\sqrt{\left(2-{\displaystyle \frac{2}{x^2}}+{\displaystyle \frac{1}{x^4}}\right)}}dx...\left(i\right)\left[\therefore \sqrt{x^4}=x^2\right] \end{gathered}$$

Substituting $t=2-\frac{2}{x^2}+\frac{1}{x^4}...\left(ii\right)$

Differentiating $t$ w.r.t. $x$

$$\begin{gathered} \Rightarrow \left(-\frac{4}{x^5}+\frac{4}{x^3}\right)dx=dt \\ \Rightarrow 4\left(\frac{x^2-1}{x^5}\right)dx=dt......\left(iii\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{1}{4}\int \frac{dt}{\sqrt{t}}=\frac{1}{4}\int t^{-\frac{1}{2}}dt[\text{from}(i),(ii)\&(iii\left)\right] \\ =\frac{1}{4}\left[\frac{t^{-\frac{1}{2}}}{{\displaystyle \frac{1}{2}}}\right]+C[C=\text{constant}] \\ =\frac{1}{2}\sqrt{t}+C....\left(iv\right) \end{gathered}$$

Substituting back $t=2-\frac{2}{x^2}+\frac{1}{x^4}$ in equation $\left(iv\right)$

$=\frac{1}{2}\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C$

Hence, option (C) is the correct option.