# Integral question

can someone show me how to do this integral:

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

## Answers and Replies

TD
Homework Helper
Integration by parts works this way, but perhaps there's an easier way.

$$\begin{array}{l} \int {\frac{{\left( {1 - x} \right)}}{{x^2 }}e^{x - 1} } dx = - \int {\left( {1 - x} \right)e^{x - 1} } d\left( {\frac{1}{x}} \right) = - \left( {\frac{{\left( {1 - x} \right)e^{x - 1} }}{x} - \int {\frac{1}{x}d\left( {\left( {1 - x} \right)e^{x - 1} } \right)} } \right) \\ \\ = - \frac{{\left( {1 - x} \right)e^{x - 1} }}{x} - \int {\frac{{ - xe^{x - 1} }}{x}dx} = - \frac{{\left( {1 - x} \right)e^{x - 1} }}{x} + e^{x - 1} + C = \frac{{e^{x - 1} }}{x} + C \\ \end{array}$$

GCT
$$I= \int \frac{(1-x)}{x^2} e^{x-1} dx$$
$$=e^{-1} \int \frac{e^{x}dx}{x^{2}} -e^{-1} \int \frac{e^{x}dx}{x}$$
$$J=\int \frac{e^{x}dx}{x^{2}} ,~K=\int \frac{e^{x}dx}{x}$$
$$K= \frac{e^{x}}{x}+ \int \frac{e^{x}dx}{x^{2}}$$
$$I=e^{-1}J- \frac{e^{-1}e^{x}}{x} -e^{-1}J,~I= \frac{-e^{-1}e^{x}}{x}+C$$