What is the integral of te^{5t} ?

The integral of te^{5t} is 1/25 (e^{5t}*5t-e^{5t})+C

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integral of te^{5t}

\int t e^{5 t} d t

\frac{1}{25} (e^{5 t} \cdot 5 t - e^{5 t}) + C

Solution steps

\int t e^{5 t} d t

Apply u-substitution

: \int \frac{e ^{u} u}{25} d u

= \int \frac{e ^{u} u}{25} d u

Take the constant out:

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

= \frac{1}{25} \cdot \int e^{u} u d u

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= \frac{1}{25} (e^{u} u - \int e^{u} d u)

\int e^{u} d u = e^{u}

= \frac{1}{25} (e^{u} u - e^{u})

Substitute back

u = 5 t

= \frac{1}{25} (e^{5 t} \cdot 5 t - e^{5 t})

Add a constant to the solution

= \frac{1}{25} (e^{5 t} \cdot 5 t - e^{5 t}) + C

x \to 2 lim (\frac{1}{2} x^{2} + 2)

\int (9 x^{2} - 5 x + 3) d x

x \to 2 lim ((x - 2)^{x - 2})

\frac{d}{d x} ((\frac{1}{2 x + 3})^{3})

f (x) = arctan (\frac{1 + x}{1 - x})

What is the integral of te^{5t} ?
The integral of te^{5t} is 1/25 (e^{5t}*5t-e^{5t})+C