The value of int frac{e^{x}left(x^{2} tan ^{ | vedclass

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

Question

(A) Given that,$ I = \int \frac{e^{x}(x^{2} 	an^{-1} x + 	an^{-1} x + 1)}{x^{2} + 1} dx $.We can rewrite the numerator by grouping terms:$ I = \int

Vedclass · Accepted Answer

$ e^{x} 	an ^{-1} x+c $

Vedclass · Answer

(A) Given that,$ I = \int \frac{e^{x}(x^{2} 	an^{-1} x + 	an^{-1} x + 1)}{x^{2} + 1} dx $.We can rewrite the numerator by grouping terms:$ I = \int e^{x} \left( \frac{(x^{2} + 1) 	an^{-1} x + 1}{x^{2} + 1} ight) dx $.Dividing each term by $ x^{2} + 1 $:$ I = \int e^{x} \left( 	an^{-1} x + \frac{1}{x^{2} + 1} ight) dx $.Let $ f(x) = 	an^{-1} x $. Then $ f'(x) = \frac{1}{x^{2} + 1} $.Using the standard integral formula $ \int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c $:$ I = e^{x} 	an^{-1} x + c $.

The value of $ \int \frac{e^{x}\left(x^{2} \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^{2}+1} d x $ is equal to

Similar Questions

$\int \left( \frac{2 - \sin 2x}{1 - \cos 2x} \right) e^x \, dx$ is equal to

The value of $\int e^{x} \left[ \frac{1+\sin x}{1+\cos x} \right] dx$ is

$\int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx = $ . . . . . . $+ c$.

$\int e^{x}\left(\frac{2}{x}-\frac{2}{x^2}\right) dx$ is equal to

$\int \frac{(x-1) e^{x}}{(x+1)^{3}} d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module