Evaluate the following integrals :\(\int\frac{\sqrt{1+x^2}}{x^4}\)dx

1.	Evaluate the following integrals :\(\int\frac{\sqrt{1+x^2}}{x^4}\)dx
Answer» Let , x = tanθ So, dx = sec 2θ dθ And, θ = tan^-x Putting above values, \(\int\frac{\sqrt{1+x^2}}{x^4}\)dx = \(\int\frac{\sqrt{1+tam^2\theta}}{tan^4\theta}\) sec²θ dθ = ∫ sec²θ/tan²θ dθ = ∫ cosec²θ dθ = - cotθ + c Put, θ = tan^-x = - cotθ + c = - cot tan^-x + c

Answer»

Let ,

x = tanθ

So,

dx = sec 2θ dθ

And,

θ = tan^-x

Putting above values,

\(\int\frac{\sqrt{1+x^2}}{x^4}\)dx

= \(\int\frac{\sqrt{1+tam^2\theta}}{tan^4\theta}\) sec²θ dθ

= ∫ sec²θ/tan²θ dθ

= ∫ cosec²θ dθ

= - cotθ + c

Put,

θ = tan^-x

= - cotθ + c

= - cot tan^-x + c

Discussion