Question

Solve:$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx$ is equal to

• A. $\frac{\sqrt{2x^4-2x^2+1}}{x^2}+C$
• B. $\frac{\sqrt{2x^4-2x^2+1}}{x^3}+C$
• C. $\frac{\sqrt{2x^4-2x^2+1}}{x}+C$
• D. $\frac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$

Answer 1

Answer: (D)

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

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

$\Rightarrow$ $\int \frac{\frac{x^2-1}{x^5}}{\frac{x^3\sqrt{2x^4-2x^2+1}}{x^5}}dx$

$\Rightarrow$ $\int \frac{\frac{1}{x^3}-\frac{1}{x^5}}{\frac{\sqrt{2x^4-2x^2+1}}{x^2}}dx$

$\Rightarrow$ $\int \frac{\frac{1}{x^3}-\frac{1}{x^5}}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}dx$

Let $2-\frac{2}{x^2}+\frac{1}{x^4}=t$

$(0+\frac{4}{x^3}-\frac{4}{x^5})dx=dt$

$(\frac{1}{x^3}-\frac{1}{x^5})=\frac{dt}{4}$

Now, substituting above value,

$\Rightarrow$ $\int \frac{dt}{4\sqrt{t}}$

$\Rightarrow$ $\frac{1}{4}\int t^{\frac{-1}{2}}dt$

$\Rightarrow$ $\frac{1}{4}\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+c$

$\Rightarrow$ $\frac{1}{4}\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c$

$\Rightarrow$ $\frac{\sqrt{t}}{2}+c$

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

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