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

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

$\int \frac{e^{\cot x}}{\sin^2 x} (2 \log \csc x + \sin 2 x) dx =$

$\int e^{4x}(\sin 3x - \cos 3x) dx = $

If $\int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx = f(x) + \text{constant}$, then $f(x)$ is equal to

Let $f(x) = \int \frac{(2-x^2)e^x}{(\sqrt{1+x})(1-x)^{3/2}} dx$. If $f(0) = 0$, then $f(\frac{1}{2})$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module