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

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

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integral of 5te^{5t}

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

e^{5 t} t - \frac{1}{5} e^{5 t} + C

Solution steps

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

Take the constant out:

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

= 5 \cdot \int t e^{5 t} d t

Apply u-substitution

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

= 5 \cdot \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

= 5 \cdot \frac{1}{25} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 5 \cdot \frac{1}{25} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = 5 t

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

Simplify

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

= e^{5 t} t - \frac{1}{5} e^{5 t}

Add a constant to the solution

= e^{5 t} t - \frac{1}{5} e^{5 t} + C

\frac{d y}{d x} = 2 (y^{2} - y) x

y = e^{(3 - x)}

\frac{\partial}{\partial z} (e^{x z} + x y)

\int \frac{e ^{5 x}}{e ^{5 x} + 9} d x

x \to - \infty lim (- x \cdot e^{x})

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