int {frac{{(x + 3){e^x}}}{{{{(x + 4)}^2}}},dx | vedclass

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

Question

(C) We need to evaluate the integral $I = \int {\frac{{(x + 3){e^x}}}{{{{(x + 4)}^2}}}dx}$.Rewrite the numerator $(x + 3)$ as $(x + 4 - 1)$:$I = \int

Vedclass · Accepted Answer

$\frac{{{e^x}}}{{x + 4}} + c$

Vedclass · Answer

(C) We need to evaluate the integral $I = \int {\frac{{(x + 3){e^x}}}{{{{(x + 4)}^2}}}dx}$.Rewrite the numerator $(x + 3)$ as $(x + 4 - 1)$:$I = \int {{e^x}\frac{{(x + 4) - 1}}{{{{(x + 4)}^2}}}dx}$$I = \int {{e^x}\left( {\frac{{x + 4}}{{{{(x + 4)}^2}}} - \frac{1}{{{{(x + 4)}^2}}}} \right)dx}$$I = \int {{e^x}\left( {\frac{1}{{x + 4}} - \frac{1}{{{{(x + 4)}^2}}}} \right)dx}$Recall the standard integral formula: $\int {{e^x}(f(x) + f'(x))dx = {e^x}f(x) + c}$.Here,let $f(x) = \frac{1}{{x + 4}}$. Then $f'(x) = -\frac{1}{{{{(x + 4)}^2}}}$.Thus,$I = {e^x}\left( {\frac{1}{{x + 4}}} \right) + c = \frac{{{e^x}}}{{x + 4}} + c$.

$\int {\frac{{(x + 3){e^x}}}{{{{(x + 4)}^2}}}\,dx} = \,$

Similar Questions

$\int [\sin(\log x) + \cos(\log x)] dx = $

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

$\int {{e^x}\left[ {f(x) + f'(x)} \right]\,dx} $ is equal to

$\int e^{\tan x}(\sec^2 x + \sec^3 x \sin x) dx =$

$\int {{e^x} \left[ \frac{1 + x \log x}{x} \right] \, dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module