What is the integral of (e^y)^3 ?

The integral of (e^y)^3 is 1/3 e^{3y}+C

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integral of (e^y)^3

integral

(e^{y})^{3}

\frac{1}{3} e^{3 y} + C

Solution steps

\int (e^{y})^{3} d y

Apply exponent rule:

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

= \int e^{3 y} d y

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Substitute back

u = 3 y

= \frac{1}{3} e^{3 y}

Add a constant to the solution

= \frac{1}{3} e^{3 y} + C

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

(sin^{2} (x)) (cos^{2} (x))

(1 + x)^{4}

cos^{7} (x)

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