What is the integral of (e^x)^4 ?

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

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integral of (e^x)^4

integral

(e^{x})^{4}

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

Solution steps

\int (e^{x})^{4} d x

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= \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

\frac{1}{( x ^{2} + x )}

\frac{x}{( 5 x + 2 ) ^{3}}

\frac{1}{( x ^{2} + 4 )}

\frac{1}{( a ^{2} - x ^{2} ) ^{\frac{1}{2}}}

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