Question

Evaluate:

${\int }_1^2\left(\frac{x-1}{x^2}\right)e^{x}dx$

Answer 1

${\int }_1^2\left(\frac{x-1}{x^2}\right)e^{x}dx$

${\int }_1^2\frac{e^{(x-1)}}{x^2}dx$

$\frac{-e^x(x-1)}{x}{|}_1^2-{\int }_1^2-e^{x}dx$ here $\Rightarrow u=e^x(x-1);v^1=\frac{1}{x^2}$

$\frac{-e^x(x-1)}{x}+e^x{|}_1^2$

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

$[\frac{-e^2}{2}+e^2-e]$

Solution of Given Definite Integral