int {{{left( {frac{{x + 2}}{{x + 4}}} right)} | vedclass

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

Question

(A) We know that $\int {{e^x}[f(x) + f'(x)]dx = {e^x}f(x) + c}$.Let $I = \int {{e^x}{{\left( {\frac{{x + 2}}{{x + 4}}} \right)}^2}dx} $.Rewrite the te

Vedclass · Accepted Answer

${e^x}\left( {\frac{x}{{x + 4}}} \right) + c$

Vedclass · Answer

(A) We know that $\int {{e^x}[f(x) + f'(x)]dx = {e^x}f(x) + c}$.Let $I = \int {{e^x}{{\left( {\frac{{x + 2}}{{x + 4}}} \right)}^2}dx} $.Rewrite the term inside the square: $\frac{{x + 2}}{{x + 4}} = \frac{{x + 4 - 2}}{{x + 4}} = 1 - \frac{2}{{x + 4}}$.So,$I = \int {{e^x}{{\left( {1 - \frac{2}{{x + 4}}} \right)}^2}dx} = \int {{e^x}\left( {1 - \frac{4}{{x + 4}} + \frac{4}{{{{(x + 4)}^2}}}} \right)dx} $.Let $f(x) = 1 - \frac{4}{{x + 4}}$. Then $f'(x) = - \frac{d}{{dx}}\left( {\frac{4}{{x + 4}}} \right) = - 4 \cdot \left( { - \frac{1}{{{{(x + 4)}^2}}}} \right) = \frac{4}{{{{(x + 4)}^2}}}$.Thus,$I = \int {{e^x}[f(x) + f'(x)]dx = {e^x}f(x) + c} $.$I = {e^x}\left( {1 - \frac{4}{{x + 4}}} \right) + c = {e^x}\left( {\frac{{x + 4 - 4}}{{x + 4}}} \right) + c = {e^x}\left( {\frac{x}{{x + 4}}} \right) + c$.

$\int {{{\left( {\frac{{x + 2}}{{x + 4}}} \right)}^2}{e^x}\,dx} $ is equal to

Similar Questions

The value of $I = \int \frac{(x-1) e^x}{(x+1)^3} \,dx$ is

If $\int e^x\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}+\frac{x}{1-x^2}\right) d x=g(x)+C$ where $C$ is the constant of integration,then $g \left(\frac{1}{2}\right)$ equals :

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

If $x \in [-1, 1]$,then the value of $\int e^{\sin^{-1} x} \left( \frac{x + \sqrt{1-x^2}}{\sqrt{1-x^2}} \right) dx$ is

The value of $\int e^x \left( \frac{x^2+4x+4}{(x+4)^2} \right) dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module