Browse Questions

# Evaluate the following $\int\frac{\sqrt{1+ x^2}}{x^4}dx$

Toolbox:
• If $f(x)$ is substituted by $f(t)$ then $f'(x)dx=f'(t)dt,$ hence $\int f(x)dx=\int f(t)dt$
• $\large\int x^n dx =\frac{x^{n+1}}{n+1}+c$
Let $\large\int\frac{\sqrt{1+ x^2}}{x^4}dx$
This can be written as
$I=\int\large\frac{\sqrt{x^2(1+\frac{1}{x^2})}}{x^4}dx$
$I=\int\large\frac{x \sqrt{(1+\frac{1}{x^2})}}{x^4}dx \qquad =\int\frac{\sqrt{(1+\frac{1}{x^2})}}{x^3}dx$
Put $1+\large\frac{1}{x^2}=t$
Differentiating with respect to t, we get,
$-2x^{-3} dx=dt => \large\frac{-2}{x^3}dx=dt$
Therefore $\large\frac{1}{x^3} dx=\frac{-dt}{2}$
substituting in I we get,
$I=\int \sqrt t . (\large\frac{-dt}{2})=\frac{-1}{2} \int \sqrt t dt$
Integrating we get
$I=\large\frac{-1}{2} \times \bigg[\frac{t^{3/2}}{3/2}\bigg] +c$
substituting for t we get,
$\large-\frac{1}{2} \times \frac{2}{3} \bigg[\bigg(1+\frac{1}{x^2}\bigg)^{3/2}\bigg]+c$
$=-\large\frac{1}{3} \bigg(1+\frac{1}{x^2}\bigg)^{3/2}+c$