What is the integral of ((t-1)e^t)/(t^2) ?

The integral of ((t-1)e^t)/(t^2) is -(e^t(t-1))/t+e^t+C

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integral of ((t-1)e^t)/(t^2)

\int \frac{( t - 1 ) e ^{t}}{t ^{2}} d t

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

Solution steps

\int \frac{( t - 1 ) e ^{t}}{t ^{2}} d t

Apply Integration By Parts

: - \frac{e ^{t} ( t - 1 )}{t} - \int - e^{t} d t

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

\int - e^{t} d t = - e^{t}

= - \frac{e ^{t} ( t - 1 )}{t} - (- e^{t})

Simplify

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

Add a constant to the solution

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

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What is the integral of ((t-1)e^t)/(t^2) ?
The integral of ((t-1)e^t)/(t^2) is -(e^t(t-1))/t+e^t+C