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

The integral of e^{3x}e^x is 1/4 e^{4x}+C

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integral of e^{3x}e^x

\int e^{3 x} e^{x} d x

\frac{1}{4} e^{4 x} + C

Solution steps

\int e^{3 x} e^{x} d x

Simplify

e^{3 x} e^{x} : e^{4 x}

= \int e^{4 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{4} d u

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{4} e^{u}

Substitute back

u = 4 x

= \frac{1}{4} e^{4 x}

Add a constant to the solution

= \frac{1}{4} e^{4 x} + C

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x \to 3 lim (x^{2} + 4)

\int_{- 1}^{1} x (1.5 x^{2}) d x

\frac{d}{d x} (ln (\frac{x + 1}{x - 1}))