int frac{1}{x^2sqrt{1+x^2}} dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int \frac{1}{x^2\sqrt{1+x^2}} dx$. Take $x^2$ common from the square root: $I = \int \frac{1}{x^2 \sqrt{x^2(1 + \frac{1}{x^2})}} dx = \i

Vedclass · Accepted Answer

$\frac{-\sqrt{x^2+1}}{x} + c$

Vedclass · Answer

(A) Let $I = \int \frac{1}{x^2\sqrt{1+x^2}} dx$. Take $x^2$ common from the square root: $I = \int \frac{1}{x^2 \sqrt{x^2(1 + \frac{1}{x^2})}} dx = \int \frac{1}{x^3 \sqrt{1 + \frac{1}{x^2}}} dx$. Let $1 + \frac{1}{x^2} = t^2$. Then,differentiating both sides with respect to $x$,we get $-\frac{2}{x^3} dx = 2t dt$,which implies $\frac{dx}{x^3} = -t dt$. Substituting these into the integral: $I = \int -t dt / t = -\int dt = -t + c$. Since $t = \sqrt{1 + \frac{1}{x^2}} = \sqrt{\frac{x^2+1}{x^2}} = \frac{\sqrt{x^2+1}}{|x|}$,for $x > 0$,$I = -\frac{\sqrt{x^2+1}}{x} + c$.

$\int \frac{1}{x^2\sqrt{1+x^2}} dx =$

Similar Questions

$ \int \frac{1}{\sqrt{x}+x \sqrt{x}} d x $

$\int \frac{d x}{e^{x}+e^{-x}}$ is equal to

$\int \tan ^8 x \cdot \sec ^4 x \, dx = $ . . . . . . $+ C$.

$\int \tan x \sec^2 x \sqrt{1 - \tan^2 x} \; dx = $

$\int e^{3 \log x}\left(x^4+1\right)^{-1} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module