int {frac{{{e^x}(x - 1)}}{{{x^2}}};dx = } | vedclass

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

Question

(A) We use the standard integral formula: $\int {e^x [f(x) + f'(x)]\,dx = e^x f(x) + c}$.Given the integral: $I = \int {\frac{{{e^x}(x - 1)}}{{{x^2}}}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We use the standard integral formula: $\int {e^x [f(x) + f'(x)]\,dx = e^x f(x) + c}$.Given the integral: $I = \int {\frac{{{e^x}(x - 1)}}{{{x^2}}}\;dx}$.Rewrite the integrand as: $I = \int {e^x \left( {\frac{x}{{{x^2}}} - \frac{1}{{{x^2}}}} \right)\,dx} = \int {e^x \left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)\,dx}$.Let $f(x) = \frac{1}{x}$. Then $f'(x) = -\frac{1}{{{x^2}}}$.Substituting this into the formula,we get: $I = e^x \left( \frac{1}{x} \right) + c = \frac{{{e^x}}}{x} + c$.

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

Similar Questions

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

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

$\int_0^1 \frac{x e^x}{(2+x)^3} d x$ is equal to

If $\int {{e^{{x^2}}}\left( {2 - \frac{1}{{{x^2}}}} \right)dx = {e^{{x^2}}}f(x) + C} $ and $f\left( {\frac{1}{2}} \right) = 2$,then $f(1)$ is equal to (where $C$ is an arbitrary constant).

$\int e^x(2021+\tan x+\tan^2 x) dx = $ . . . . . . $+ C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module