What is the integral of t^2e^{5t} ?

The integral of t^2e^{5t} is 1/5 e^{5t}t^2-2/125 (5e^{5t}t-e^{5t})+C

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integral of t^2e^{5t}

\int t^{2} e^{5 t} d t

\frac{1}{5} e^{5 t} t^{2} - \frac{2}{125} (5 e^{5 t} t - e^{5 t}) + C

Solution steps

\int t^{2} e^{5 t} d t

Apply Integration By Parts

= \frac{1}{5} e^{5 t} t^{2} - \int \frac{2}{5} e^{5 t} t d t

\int \frac{2}{5} e^{5 t} t d t = \frac{2}{125} (5 e^{5 t} t - e^{5 t})

= \frac{1}{5} e^{5 t} t^{2} - \frac{2}{125} (5 e^{5 t} t - e^{5 t})

Add a constant to the solution

= \frac{1}{5} e^{5 t} t^{2} - \frac{2}{125} (5 e^{5 t} t - e^{5 t}) + C

\frac{d}{d x} (e^{2 x^{2} - 5 x})

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

in v erse l a pl a ce \frac{2}{s ^{2} + 4}

\int \frac{5 - 3 sin ( 2 x )}{sec ( 6 x )} d x

\int (\frac{x ^{2} + 1}{x}) d x

What is the integral of t^2e^{5t} ?
The integral of t^2e^{5t} is 1/5 e^{5t}t^2-2/125 (5e^{5t}t-e^{5t})+C