What is the integral of 9xe^{5x} ?

The integral of 9xe^{5x} is 9/25 (5e^{5x}x-e^{5x})+C

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integral of 9xe^{5x}

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

\frac{9}{25} (5 e^{5 x} x - e^{5 x}) + C

Solution steps

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

Apply u-substitution

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

= 9 \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

= 9 \cdot \frac{1}{25} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 9 \cdot \frac{1}{25} (e^{u} u - \int e^{u} d u)

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

= 9 \cdot \frac{1}{25} (e^{u} u - e^{u})

Substitute back

u = 5 x

= 9 \cdot \frac{1}{25} (e^{5 x} \cdot 5 x - e^{5 x})

Simplify

9 \cdot \frac{1}{25} (e^{5 x} \cdot 5 x - e^{5 x}) : \frac{9}{25} (5 e^{5 x} x - e^{5 x})

= \frac{9}{25} (5 e^{5 x} x - e^{5 x})

Add a constant to the solution

= \frac{9}{25} (5 e^{5 x} x - e^{5 x}) + C

6 x, 0, 2

x^{2} + 2 x + 1, - 3, 0

f (x) = - 4 x^{2} - 3 x + 2, at x = - 1

\int_{1}^{3} x^{2} \frac{2.25}{x ^{3}} d x

\int (3 x + 1)^{5} d x

What is the integral of 9xe^{5x} ?
The integral of 9xe^{5x} is 9/25 (5e^{5x}x-e^{5x})+C