What is the integral of 6xe^{5x} ?

The integral of 6xe^{5x} is 6/25 (5e^{5x}x-e^{5x})+C

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integral of 6xe^{5x}

\int 6 x e^{5 x} d x

\frac{6}{25} (5 e^{5 x} x - e^{5 x}) + C

Solution steps

\int 6 x e^{5 x} d x

Take the constant out:

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

= 6 \cdot \int x e^{5 x} d x

Apply u-substitution

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

= 6 \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

= 6 \cdot \frac{1}{25} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 6 \cdot \frac{1}{25} (e^{u} u - \int e^{u} d u)

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

= 6 \cdot \frac{1}{25} (e^{u} u - e^{u})

Substitute back

u = 5 x

= 6 \cdot \frac{1}{25} (e^{5 x} \cdot 5 x - e^{5 x})

Simplify

6 \cdot \frac{1}{25} (e^{5 x} \cdot 5 x - e^{5 x}) : \frac{6}{25} (5 e^{5 x} x - e^{5 x})

= \frac{6}{25} (5 e^{5 x} x - e^{5 x})

Add a constant to the solution

= \frac{6}{25} (5 e^{5 x} x - e^{5 x}) + C

\int x^{2} (5 x^{3} - 1)^{4} d x

\int \frac{x + 1}{x ^{2} - 16} d x

n = 0 \sum \infty 4 n^{2} + 4 n + 1

y = 4 x^{2}, y = x^{2} + 2

x \to 0 lim ((1 - 4 x)^{\frac{1}{x}})

What is the integral of 6xe^{5x} ?
The integral of 6xe^{5x} is 6/25 (5e^{5x}x-e^{5x})+C