What is the integral of t^{-4}e^t ?

The integral of t^{-4}e^t is -(e^t}{3t^3}+1/3 (-(e^t)/(2t^2)+1/2 (-\frac{e^t)/t+\Ei(t)))+C

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integral of t^{-4}e^t

\int t^{- 4} e^{t} d t

- \frac{e ^{t}}{3 t ^{3}} + \frac{1}{3} (- \frac{e ^{t}}{2 t ^{2}} + \frac{1}{2} (- \frac{e ^{t}}{t} + Ei (t))) + C

Solution steps

\int t^{- 4} e^{t} d t

Apply Integration By Parts

: - \frac{e ^{t}}{3 t ^{3}} - \int - \frac{e ^{t}}{3 t ^{3}} d t

= - \frac{e ^{t}}{3 t ^{3}} - \int - \frac{e ^{t}}{3 t ^{3}} d t

\int - \frac{e ^{t}}{3 t ^{3}} d t = - \frac{1}{3} (- \frac{e ^{t}}{2 t ^{2}} + \frac{1}{2} (- \frac{e ^{t}}{t} + Ei (t)))

= - \frac{e ^{t}}{3 t ^{3}} - (- \frac{1}{3} (- \frac{e ^{t}}{2 t ^{2}} + \frac{1}{2} (- \frac{e ^{t}}{t} + Ei (t))))

Simplify

= - \frac{e ^{t}}{3 t ^{3}} + \frac{1}{3} (- \frac{e ^{t}}{2 t ^{2}} + \frac{1}{2} (- \frac{e ^{t}}{t} + Ei (t)))

Add a constant to the solution

= - \frac{e ^{t}}{3 t ^{3}} + \frac{1}{3} (- \frac{e ^{t}}{2 t ^{2}} + \frac{1}{2} (- \frac{e ^{t}}{t} + Ei (t))) + C

\frac{d}{d x} (7 x^{3} + 3 x - 12)

\frac{d}{d x} (\frac{x}{4 x + 3 ln ( x )})

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\frac{\partial}{\partial x} (3 x y z^{2})

s + 1

What is the integral of t^{-4}e^t ?
The integral of t^{-4}e^t is -(e^t}{3t^3}+1/3 (-(e^t)/(2t^2)+1/2 (-\frac{e^t)/t+\Ei(t)))+C