Question

$\int \frac{\sqrt{5+x^2}}{x^4}dx$ is equal to

• A. $\frac{1}{15}{(1+\frac{5}{x^2})}^{\frac{3}{2}}+C$
• B. $\frac{-1}{15}{(1+\frac{1}{x^2})}^{\frac{3}{2}}+C$
• C. $\frac{-1}{15}{(1+\frac{5}{x^2})}^{\frac{3}{2}}+C$
• D. $\frac{1}{15}{(1+\frac{1}{x^2})}^{\frac{3}{2}}+C$
• E. $\frac{-1}{10}{(1+\frac{1}{x^2})}^{\frac{3}{2}}+C$

Answer 1

The correct option is C $\frac{-1}{15}{(1+\frac{5}{x^2})}^{\frac{3}{2}}+C$

Consider, $I=\int \frac{\sqrt{5+x^2}}{x^4}dx$

$I=\int \frac{\sqrt{\frac{5}{x^2}+1}}{\left(x\right)\left(x^3\right)}dx$

Put $t=\frac{5}{x^2}+1$

$\frac{dt}{dx}=\frac{-10}{x^3}$ $\Rightarrow -\frac{dt}{10}=\frac{dx}{x^3}$

Substituting this in the given integral, we get

$\Rightarrow -\frac{dt}{10}=\frac{dx}{x^3}\Rightarrow -\frac{1}{10}\int \sqrt{t}dt\Rightarrow \frac{-1}{10}\frac{t^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+c\Rightarrow -\frac{t^{\frac{3}{2}}}{15}+c\Rightarrow -\frac{1}{15}{(\frac{5}{x^2}+1)}^{\frac{3}{2}}+c$

So, option C is correct.