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

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

Question

(A) To evaluate the integral $I = \int \frac{\sqrt{x}}{1+x} dx$,substitute $x = t^2$,so $dx = 2t dt$.Substituting these into the integral:$I = \int \f

Vedclass · Accepted Answer

$2(\sqrt{x} - 	an^{-1}\sqrt{x}) + c$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{\sqrt{x}}{1+x} dx$,substitute $x = t^2$,so $dx = 2t dt$.Substituting these into the integral:$I = \int \frac{t}{1+t^2} (2t dt) = 2 \int \frac{t^2}{1+t^2} dt$$I = 2 \int \frac{t^2+1-1}{1+t^2} dt = 2 \int \left(1 - \frac{1}{1+t^2}ight) dt$$I = 2 \left( \int 1 dt - \int \frac{1}{1+t^2} dt ight)$$I = 2(t - 	an^{-1}t) + c$Substituting $t = \sqrt{x}$ back:$I = 2(\sqrt{x} - 	an^{-1}\sqrt{x}) + c$.

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

Similar Questions

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

Find the value of $k$ if $\int \cos ^k(x) \sin (x) d x = \frac{-1}{4} \cos ^4(x) + C$.

$\int \frac{(3 x-2) \tan \left(\sqrt{9 x^2-12 x+1}\right)}{\sqrt{9 x^2-12 x+1}} d x=$

$\int \sin^3 x \cos^2 x \, dx = $

Integrate the function $\sin x \cdot \sin (\cos x)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module