int frac{x^{49} tan ^{-1}left(x^{50}right)}{1 | vedclass

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

Question

(C) Let $I = \int \frac{x^{49} 	an ^{-1}\left(x^{50}ight)}{1+x^{100}} d x$.Substitute $t = x^{50}$,then $dt = 50x^{49} dx$,which implies $x^{49} dx

Vedclass · Accepted Answer

$\frac{1}{100}$

Vedclass · Answer

(C) Let $I = \int \frac{x^{49} 	an ^{-1}\left(x^{50}ight)}{1+x^{100}} d x$.Substitute $t = x^{50}$,then $dt = 50x^{49} dx$,which implies $x^{49} dx = \frac{1}{50} dt$.The integral becomes $I = \frac{1}{50} \int \frac{	an ^{-1} t}{1+t^2} dt$.Now,substitute $u = 	an ^{-1} t$,then $du = \frac{1}{1+t^2} dt$.Substituting this into the integral,we get $I = \frac{1}{50} \int u du = \frac{1}{50} \cdot \frac{u^2}{2} + c = \frac{u^2}{100} + c$.Replacing $u$ back,we get $I = \frac{(	an ^{-1} x^{50})^2}{100} + c$.Comparing this with the given expression $k(	an ^{-1} x^{50})^2 + c$,we find $k = \frac{1}{100}$.

$\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{1+x^{100}} d x = k\left(\tan ^{-1}\left(x^{50}\right)\right)^2 + c$,then $k$ is equal to

Similar Questions

$A$ circle of radius $4$,drawn on a chord of the parabola $y^2 = 8x$ as diameter,touches the axis of the parabola. Then,the slope of the chord is

The decreasing order of bond dissociation energies of $C-C$,$C-H$,and $H-H$ bonds is:

In a Ruby laser, the colour of laser light is due to ______ atom.

The plant part which consists of two generations,one within the other,is:

The maximum value of $\left(\frac{1}{x}\right)^{x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module