int xsqrt{1 + x^2} , dx = | vedclass

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

Question

(D) To evaluate the integral $\int x\sqrt{1 + x^2} \, dx$,we use the method of substitution.Let $t = 1 + x^2$.Then,differentiating both sides with res

Vedclass · Accepted Answer

$\frac{1}{3}(1 + x^2)^{3/2} + c$

Vedclass · Answer

(D) To evaluate the integral $\int x\sqrt{1 + x^2} \, dx$,we use the method of substitution.Let $t = 1 + x^2$.Then,differentiating both sides with respect to $x$,we get $dt = 2x \, dx$,which implies $x \, dx = \frac{dt}{2}$.Substituting these into the integral,we get:$\int \sqrt{t} \cdot \frac{dt}{2} = \frac{1}{2} \int t^{1/2} \, dt$.Using the power rule for integration,$\int t^n \, dt = \frac{t^{n+1}}{n+1} + c$,we have:$\frac{1}{2} \cdot \frac{t^{3/2}}{3/2} + c = \frac{1}{2} \cdot \frac{2}{3} t^{3/2} + c = \frac{1}{3} t^{3/2} + c$.Finally,substituting $t = 1 + x^2$ back,we get $\frac{1}{3}(1 + x^2)^{3/2} + c$.

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

Similar Questions

$\int (1+e^{-x})^{-1} dx =$

$\int \frac{(x^4+x)^{\frac{1}{4}}}{x^5} dx = $ . . . . . . $+ C$.

Let $n \ge 2$ be a natural number and $0 < \theta < \frac{\pi}{2}$. Then $\int \frac{(\sin^n \theta - \sin \theta)^{\frac{1}{n}} \cos \theta}{\sin^{n+1} \theta} d\theta$ is equal to

The integral $\int \sec^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x \, dx$ is equal to

$\int \frac{x(x \sin x+\cos x)^{-2}}{\sec x} d x=$ . . . . . . $+C$

Vedclass Test Series

Exam Paper Generator

Online Exam Module