What is the integral of 5e^{3x+7} ?

The integral of 5e^{3x+7} is 5/3 e^{3x+7}+C

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integral of 5e^{3x+7}

\int 5 e^{3 x + 7} d x

\frac{5}{3} e^{3 x + 7} + C

Solution steps

\int 5 e^{3 x + 7} d x

Take the constant out:

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

= 5 \cdot \int e^{3 x + 7} d x

Apply u-substitution

: \int e^{u} \frac{1}{3} d u

= 5 \cdot \int e^{u} \frac{1}{3} d u

Take the constant out:

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

= 5 \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= 5 \cdot \frac{1}{3} e^{u}

Substitute back

u = 3 x + 7

= 5 \cdot \frac{1}{3} e^{3 x + 7}

Simplify

5 \cdot \frac{1}{3} e^{3 x + 7} : \frac{5}{3} e^{3 x + 7}

= \frac{5}{3} e^{3 x + 7}

Add a constant to the solution

= \frac{5}{3} e^{3 x + 7} + C

\int_{0}^{5} (x^{2} - 5 x) d x

n = 0 \sum \infty (- 1)^{n} 2

\frac{21}{x ^{3}}, 1, 1000

x y^{^{'}} - y = 2 x^{2} - x + 1

y^{^{''}} + 16 y = 3 sec (4 t)