What is the integral of 8te^{4t} ?

The integral of 8te^{4t} is 1/2 (4e^{4t}t-e^{4t})+C

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integral of 8te^{4t}

\int 8 t e^{4 t} d t

\frac{1}{2} (4 e^{4 t} t - e^{4 t}) + C

Solution steps

\int 8 t e^{4 t} d t

Take the constant out:

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

= 8 \cdot \int t e^{4 t} d t

Apply u-substitution

: \int \frac{e ^{u} u}{16} d u

= 8 \cdot \int \frac{e ^{u} u}{16} d u

Take the constant out:

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

= 8 \cdot \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 8 \cdot \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

= 8 \cdot \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = 4 t

= 8 \cdot \frac{1}{16} (e^{4 t} \cdot 4 t - e^{4 t})

Simplify

8 \cdot \frac{1}{16} (e^{4 t} \cdot 4 t - e^{4 t}) : \frac{1}{2} (4 e^{4 t} t - e^{4 t})

= \frac{1}{2} (4 e^{4 t} t - e^{4 t})

Add a constant to the solution

= \frac{1}{2} (4 e^{4 t} t - e^{4 t}) + C

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What is the integral of 8te^{4t} ?
The integral of 8te^{4t} is 1/2 (4e^{4t}t-e^{4t})+C