What is the integral of 8te^{7t} ?

The integral of 8te^{7t} is 8/49 (7e^{7t}t-e^{7t})+C

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integral of 8te^{7t}

\int 8 t e^{7 t} d t

\frac{8}{49} (7 e^{7 t} t - e^{7 t}) + C

Solution steps

\int 8 t e^{7 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^{7 t} d t

Apply u-substitution

: \int \frac{e ^{u} u}{49} d u

= 8 \cdot \int \frac{e ^{u} u}{49} d u

Take the constant out:

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

= 8 \cdot \frac{1}{49} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 8 \cdot \frac{1}{49} (e^{u} u - \int e^{u} d u)

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

= 8 \cdot \frac{1}{49} (e^{u} u - e^{u})

Substitute back

u = 7 t

= 8 \cdot \frac{1}{49} (e^{7 t} \cdot 7 t - e^{7 t})

Simplify

8 \cdot \frac{1}{49} (e^{7 t} \cdot 7 t - e^{7 t}) : \frac{8}{49} (7 e^{7 t} t - e^{7 t})

= \frac{8}{49} (7 e^{7 t} t - e^{7 t})

Add a constant to the solution

= \frac{8}{49} (7 e^{7 t} t - e^{7 t}) + C

y = x^{3 - 5 x^{2}} + 7 x + 3

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

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

- 3 x^{2} - 10 x, 3

\int \frac{3 x ^{2}}{x} d x

What is the integral of 8te^{7t} ?
The integral of 8te^{7t} is 8/49 (7e^{7t}t-e^{7t})+C