What is the integral of 3y^2(e^{y^3}-1) ?

The integral of 3y^2(e^{y^3}-1) is e^{y^3}-y^3+C

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integral of 3y^2(e^{y^3}-1)

\int 3 y^{2} (e^{y^{3}} - 1) d y

e^{y^{3}} - y^{3} + C

Solution steps

\int 3 y^{2} (e^{y^{3}} - 1) d y

Take the constant out:

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

= 3 \cdot \int y^{2} (e^{y^{3}} - 1) d y

Apply u-substitution

: \int \frac{u ( e ^{u^{\frac{3}{2}}} - 1 )}{2 y} d u

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

Take the constant out:

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d u = u

= 3 \cdot \frac{1}{3} (e^{u} - u)

Substitute back

u = y^{3}

= 3 \cdot \frac{1}{3} (e^{y^{3}} - y^{3})

Simplify

3 \cdot \frac{1}{3} (e^{y^{3}} - y^{3}) : e^{y^{3}} - y^{3}

= e^{y^{3}} - y^{3}

Add a constant to the solution

= e^{y^{3}} - y^{3} + C

n = 1001 \sum \infty \frac{1}{n}

\int_{0}^{5} x^{3} x^{2} + 25 d x

x y^{^{'}} + (11 + x) y = e^{- x}

\int \frac{x ^{3}}{x ^{2} + 36} d x

\frac{d}{d x} (- \frac{1}{2 ( x + 2 ) ^{\frac{3}{2}}})

What is the integral of 3y^2(e^{y^3}-1) ?
The integral of 3y^2(e^{y^3}-1) is e^{y^3}-y^3+C