Question

Evaluate: $\int \frac{(x^2-1)dx}{x^3\left(\sqrt{2x^4-2x^2+1}\right)}$

• A. $\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+c$
• B. $2\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+c$
• 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 $\frac{1}{2}\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+c$

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

Substitute $x^2=u\Rightarrow dx=\frac{1}{2x}du$

$=\frac{1}{2}\int \frac{u-1}{u^2\sqrt{2(u^2-u)+1}}dx$

Completing the square

$\frac{1}{2}\int \frac{u-1}{u^2\sqrt{{(\sqrt{2}u-\frac{1}{\sqrt{2}})}^2+\frac{1}{2}}}du$

$=\frac{\sqrt{2}}{2}\int \frac{u-1}{u^2\sqrt{(2u-1{)}^2+1}}du$

Substitute $v=2u-1\Rightarrow dv=2du$

$=\frac{1}{\sqrt{2}}\int \frac{v-1}{(v+1{)}^2\sqrt{v^2+1}}dv$

$=\frac{1}{\sqrt{2}}[-\frac{2(\frac{\sqrt{v^2+1}-1}{v}+1)}{\frac{(\sqrt{v^2+1}-1{)}^2}{v^2}-2\left(\frac{\sqrt{v^2+1}-1}{v}\right)-1}]$

undo substitute $v=2u-1$ & $u=x^2$

$=\left[\frac{\sqrt{2}(\frac{\sqrt{(2x^2-1{)}^2+1-1}}{2x^2-1}+1)}{\frac{(\sqrt{(2x^2-1{)}^2+1-1}{)}^2}{(2x^2-1{)}^2}-\frac{2\sqrt{(2x^1-1{)}^2+1)}}{2x^2-1}-1}\right]+C$

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

Rewriting above equation

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

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