Question

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

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

Answer 1

The correct option is C

$\left(-\frac{1}{15}\right){\left[1+\left(\frac{5}{x^2}\right)\right]}^{\frac{3}{2}}+c$

Evaluate the integral $\int \left\{\frac{\sqrt{\left(5+x^2\right)}}{x^4}\right\}dx$

Consider the given Equation as

$I=\int \left\{\frac{\sqrt{\left(5+x^2\right)}}{x^4}\right\}dx$

Rewrite the above Equation as

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

Let us assume that

$$\begin{gathered} t=\frac{5}{x^2}+1 \\ \frac{dt}{dx}=-\frac{2\times 5}{x^3} \\ -\frac{dt}{10}=\frac{dx}{x^3} \end{gathered}$$

Substitute the above values in the integral,

$$\begin{gathered} I=-\frac{1}{10}\int \sqrt{t}dt \\ I=-\frac{1}{10}\frac{t^{{\displaystyle \frac{3}{2}}}}{\left({\displaystyle \frac{3}{2}}\right)}+c \\ I=-\frac{t^{\frac{3}{2}}}{15} \\ I=-\frac{1}{15}{\left(\frac{5}{x^2}+1\right)}^{\frac{3}{2}}+c \end{gathered}$$

Hence, the correct answer is Option C.