What is the integral of 4xe^{5x} ?

The integral of 4xe^{5x} is 4/25 (5e^{5x}x-e^{5x})+C

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integral of 4xe^{5x}

\int 4 x e^{5 x} d x

\frac{4}{25} (5 e^{5 x} x - e^{5 x}) + C

Solution steps

\int 4 x e^{5 x} d x

Take the constant out:

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

= 4 \cdot \int x e^{5 x} d x

Apply u-substitution

= 4 \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

= 4 \cdot \frac{1}{25} \cdot \int e^{u} u d u

Apply Integration By Parts

= 4 \cdot \frac{1}{25} (e^{u} u - \int e^{u} d u)

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

= 4 \cdot \frac{1}{25} (e^{u} u - e^{u})

Substitute back

u = 5 x

= 4 \cdot \frac{1}{25} (e^{5 x} \cdot 5 x - e^{5 x})

Simplify

4 \cdot \frac{1}{25} (e^{5 x} \cdot 5 x - e^{5 x}) : \frac{4}{25} (5 e^{5 x} x - e^{5 x})

= \frac{4}{25} (5 e^{5 x} x - e^{5 x})

Add a constant to the solution

= \frac{4}{25} (5 e^{5 x} x - e^{5 x}) + C

(0, - 5), (- 2, 1)

f (x) = \frac{2 x}{x + 3}

(x^{2} + e^{x}) (x)

n = 1 \sum \infty (- 0.3)^{n - 1}

x \to 3 + lim (\frac{5 x}{x ^{2} - 9})

What is the integral of 4xe^{5x} ?
The integral of 4xe^{5x} is 4/25 (5e^{5x}x-e^{5x})+C