Question

Evaluate: $\int \frac{dx}{x^4\sqrt{1+x^2}}$

• A. $\frac{(2x^2+1)\sqrt{1+x^2}}{3x^3}+c$
• B. $\frac{(x^2-1)\sqrt{1-x^2}}{3x^3}+c$
• C. $\frac{(2x^2-1)\sqrt{1+x^2}}{3x^3}+c$
• D. None of these

Answer 1

The correct option is C $\frac{(2x^2-1)\sqrt{1+x^2}}{3x^3}+c$

Given : $\int \frac{dx}{x^4\sqrt{1+x^2}}=\int \frac{dx}{x^5\sqrt{\frac{1}{x^2}+1}}$

Let : $\frac{1}{x^2}+1=z^2$

$\Rightarrow -\frac{2}{x^3}dx=2zdz\Rightarrow \frac{dx}{x^3}=-zdz$

$\int \frac{dx}{x^3{x}^2\sqrt{\frac{1}{x^2}+1}}=\int -\frac{z(z^2-1)dz}{z}=\int (1-z^2)dz=z-\frac{z^3}{3}$

$=\sqrt{\frac{1}{x^2}+1}-\frac{1}{3}{\left(\sqrt{\frac{1}{x^2}+1}\right)}^2$

$=\frac{1}{3}\sqrt{\frac{1}{x^2}+1}|3-(1+\frac{1}{x^2})|$

$=\frac{\sqrt{x^2+1}}{3x}\left(\frac{2x^2-1}{x^2}\right)$

$=\frac{(2x^2-1)\sqrt{x^2+1}}{3x^3}+C$