int frac{(x-3) e^x}{(x-1)^3} d x= . . . . . | vedclass

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

Question

(D) We want to evaluate the integral $I = \int \frac{(x-3) e^x}{(x-1)^3} d x$. Rewrite the numerator as $(x-1-2)$: $I = \int \frac{(x-1-2) e^x}{(x-1)^

Vedclass · Accepted Answer

$\frac{e^x}{(x-1)^2}$

Vedclass · Answer

(D) We want to evaluate the integral $I = \int \frac{(x-3) e^x}{(x-1)^3} d x$. Rewrite the numerator as $(x-1-2)$: $I = \int \frac{(x-1-2) e^x}{(x-1)^3} d x$ $I = \int \left( \frac{x-1}{(x-1)^3} - \frac{2}{(x-1)^3} \right) e^x d x$ $I = \int \left( \frac{1}{(x-1)^2} - \frac{2}{(x-1)^3} \right) e^x d x$ Recall the standard integral form $\int e^x [f(x) + f'(x)] d x = e^x f(x) + C$. Let $f(x) = \frac{1}{(x-1)^2} = (x-1)^{-2}$. Then $f'(x) = -2(x-1)^{-3} = -\frac{2}{(x-1)^3}$. Since the integrand is in the form $e^x [f(x) + f'(x)]$,the integral is $e^x f(x) + C$. Therefore,$I = \frac{e^x}{(x-1)^2} + C$.

$\int \frac{(x-3) e^x}{(x-1)^3} d x=$ . . . . . . $+C$.

Similar Questions

$\int {e^x \frac{x^2 + 1}{(x + 1)^2} dx} = $

$\int \frac{\log _e x}{\left(1+\log _e x\right)^2} d x=$

Integrate the function: $e^{x} \left( \frac{1+\sin x}{1+\cos x} \right)$

Evaluate the integral: $\int {\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{(1 + {x^2})}}\,\,\left[ {{{\left( {{{\sec }^{ - 1}}\,\sqrt {1 + {x^2}} } \right)}^2}\,\, + \,\,{{\cos }^{ - 1}}\,\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)} \right]} \,\,\,dx$ for $x > 0$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module