\(\rm \displaystyle\int \dfrac{x^2 - 1}{x^3 \sqrt{2x^4 - 2x^2 + 1}}dx\

1.	\(\rm \displaystyle\int \dfrac{x^2 - 1}{x^3 \sqrt{2x^4 - 2x^2 + 1}}dx\) is equal to1. \(\rm \dfrac{\sqrt{2x^4 - 2x^2 + 1}}{x^2}+C\)2. \(\rm \dfrac{\sqrt{2x^4 - 2x^2 + 1}}{x^3}+C\)3. \(\rm \dfrac{\sqrt{2x^4 - 2x^2 + 1}}{x}+C\)4. \(\rm \dfrac{\sqrt{2x^4 - 2x^2 + 1}}{2x^2}+C\)
Answer» Correct Answer - Option 4 : \(\rm \dfrac{\sqrt{2x^4 - 2x^2 + 1}}{2x^2}+C\) Concept: \(\rm \int x^ndx = \frac{x^{n+1}}{n+1}+c\) Calculation: I = \(\rm \int \frac{x^2 - 1}{x^3 \sqrt{2x^4 - 2x^2 + 1}}dx\) \(=\rm \frac{1}{4}\int \frac{4x^2 - 4}{x^3 \sqrt{x^4 \left(2 - \frac{2}{x^2} + \frac{1}{x^4} \right )}}dx\) \(=\rm \frac{1}{4}\int \frac{4x^2 - 4}{x^5 \sqrt{\left(2 - \frac{2}{x^2} + \frac{1}{x^4} \right )}}dx\) \(=\rm \frac{1}{4}\int \frac{\frac{4}{x^3} - \frac{4}{x^5 }}{\sqrt{\left(2 - \frac{2}{x^2} + \frac{1}{x^4} \right )}}dx\) Let \(\rm 2 - \frac{2}{x^2} + \frac{1}{x^4} = t\) Differentiating with respect to x, we get \(\rm \left(\frac{4}{x^3} - \frac{4}{x^5 }\right)dx = dt\) Now, I \(=\rm \frac{1}{4}\int \frac{dt}{\sqrt t}\) \(=\rm \frac{1}{4}\int t^{-1/2}\;dt\) \(=\rm \frac{1}{4}\times \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\) \(=\rm \frac{1}{2}t^{\frac{1}{2}}+c\) \(=\rm \frac{1}{2} \sqrt {\rm 2 - \frac{2}{x^2} + \frac{1}{x^4}}+c\) \(=\rm \frac{\sqrt{2x^4 - 2x^2 + 1}}{2x^2}+C\)

\(\rm \displaystyle\int \dfrac{x^2 - 1}{x^3 \sqrt{2x^4 - 2x^2 + 1}}dx\) is equal to1. \(\rm \dfrac{\sqrt{2x^4 - 2x^2 + 1}}{x^2}+C\)2. \(\rm \dfrac{\sqrt{2x^4 - 2x^2 + 1}}{x^3}+C\)3. \(\rm \dfrac{\sqrt{2x^4 - 2x^2 + 1}}{x}+C\)4. \(\rm \dfrac{\sqrt{2x^4 - 2x^2 + 1}}{2x^2}+C\)

Answer» Correct Answer - Option 4 : \(\rm \dfrac{\sqrt{2x^4 - 2x^2 + 1}}{2x^2}+C\)

Concept:

\(\rm \int x^ndx = \frac{x^{n+1}}{n+1}+c\)

Calculation:

I = \(\rm \int \frac{x^2 - 1}{x^3 \sqrt{2x^4 - 2x^2 + 1}}dx\)

\(=\rm \frac{1}{4}\int \frac{4x^2 - 4}{x^3 \sqrt{x^4 \left(2 - \frac{2}{x^2} + \frac{1}{x^4} \right )}}dx\)

\(=\rm \frac{1}{4}\int \frac{4x^2 - 4}{x^5 \sqrt{\left(2 - \frac{2}{x^2} + \frac{1}{x^4} \right )}}dx\)

\(=\rm \frac{1}{4}\int \frac{\frac{4}{x^3} - \frac{4}{x^5 }}{\sqrt{\left(2 - \frac{2}{x^2} + \frac{1}{x^4} \right )}}dx\)

Let \(\rm 2 - \frac{2}{x^2} + \frac{1}{x^4} = t\)

Differentiating with respect to x, we get

\(\rm \left(\frac{4}{x^3} - \frac{4}{x^5 }\right)dx = dt\)

Now,

I \(=\rm \frac{1}{4}\int \frac{dt}{\sqrt t}\)

\(=\rm \frac{1}{4}\int t^{-1/2}\;dt\)

\(=\rm \frac{1}{4}\times \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\)

\(=\rm \frac{1}{2}t^{\frac{1}{2}}+c\)

\(=\rm \frac{1}{2} \sqrt {\rm 2 - \frac{2}{x^2} + \frac{1}{x^4}}+c\)

\(=\rm \frac{\sqrt{2x^4 - 2x^2 + 1}}{2x^2}+C\)