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

The integral of 1/(t^2)*(e^{-1})/t is -1/(2et^2)+C

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

integral

\frac{1}{t ^{2}} \cdot \frac{e ^{- 1}}{t}

- \frac{1}{2 e t ^{2}} + C

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Simplify

\frac{1}{t ^{2}} \cdot \frac{1}{t} : \frac{1}{t ^{3}}

= \frac{1}{e} \cdot \int \frac{1}{t ^{3}} d t

Apply the Power Rule

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

Simplify

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

= - \frac{1}{2 e t ^{2}}

Add a constant to the solution

= - \frac{1}{2 e t ^{2}} + C

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