What is the integral of (e^{x-1})/(x^2) ?

The integral of (e^{x-1})/(x^2) is 1/e (-(e^x)/x+\Ei(x))+C

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integral of (e^{x-1})/(x^2)

\int \frac{e ^{x - 1}}{x ^{2}} d x

\frac{1}{e} (- \frac{e ^{x}}{x} + Ei (x)) + C

Solution steps

\int \frac{e ^{x - 1}}{x ^{2}} d x

Simplify

e^{x - 1} : e^{x} e^{- 1}

= \int \frac{e ^{x} e ^{- 1}}{x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= e^{- 1} \cdot \int \frac{e ^{x}}{x ^{2}} d x

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{e} \cdot \int \frac{e ^{x}}{x ^{2}} d x

Apply Integration By Parts

: - \frac{e ^{x}}{x} - \int - \frac{e ^{x}}{x} d x

= \frac{1}{e} (- \frac{e ^{x}}{x} - \int - \frac{e ^{x}}{x} d x)

\int - \frac{e ^{x}}{x} d x = - Ei (x)

= \frac{1}{e} (- \frac{e ^{x}}{x} - (- Ei (x)))

Simplify

= \frac{1}{e} (- \frac{e ^{x}}{x} + Ei (x))

Add a constant to the solution

= \frac{1}{e} (- \frac{e ^{x}}{x} + Ei (x)) + C

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What is the integral of (e^{x-1})/(x^2) ?
The integral of (e^{x-1})/(x^2) is 1/e (-(e^x)/x+\Ei(x))+C