What is the integral of te^{at} ?

The integral of te^{at} is e^{at} 1/a t-e^{at} 1/(a^2)+C

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integral of te^{at}

\int t e^{a t} d t

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

Solution steps

\int t e^{a t} d t

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

n = 0 \sum \infty \frac{8}{n} - \frac{8}{n + 1}

5^{^{'}}

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

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

lo g_{9} (ln (x))

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