What is the integral of 9xe^{7x} ?

The integral of 9xe^{7x} is 9/49 (7e^{7x}x-e^{7x})+C

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integral of 9xe^{7x}

\int 9 x e^{7 x} d x

\frac{9}{49} (7 e^{7 x} x - e^{7 x}) + C

Solution steps

\int 9 x e^{7 x} d x

Take the constant out:

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

= 9 \cdot \int x e^{7 x} d x

Apply u-substitution

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

= 9 \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

= 9 \cdot \frac{1}{49} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 9 \cdot \frac{1}{49} (e^{u} u - \int e^{u} d u)

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

= 9 \cdot \frac{1}{49} (e^{u} u - e^{u})

Substitute back

u = 7 x

= 9 \cdot \frac{1}{49} (e^{7 x} \cdot 7 x - e^{7 x})

Simplify

9 \cdot \frac{1}{49} (e^{7 x} \cdot 7 x - e^{7 x}) : \frac{9}{49} (7 e^{7 x} x - e^{7 x})

= \frac{9}{49} (7 e^{7 x} x - e^{7 x})

Add a constant to the solution

= \frac{9}{49} (7 e^{7 x} x - e^{7 x}) + C

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What is the integral of 9xe^{7x} ?
The integral of 9xe^{7x} is 9/49 (7e^{7x}x-e^{7x})+C