What is the integral of-9x^3e^{x^4} ?

The integral of-9x^3e^{x^4} is -9/4 e^{x^4}+C

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integral of-9x^3e^{x^4}

\int - 9 x^{3} e^{x^{4}} d x

- \frac{9}{4} e^{x^{4}} + C

Solution steps

\int - 9 x^{3} e^{x^{4}} d x

Take the constant out:

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

= - 9 \cdot \int x^{3} e^{x^{4}} d x

Apply u-substitution

: \int \frac{e ^{u^{\frac{4}{3}}} u ^{\frac{1}{3}}}{3} d u

= - 9 \cdot \int \frac{e ^{u^{\frac{4}{3}}} u ^{\frac{1}{3}}}{3} d u

Take the constant out:

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

= - 9 \cdot \frac{1}{3} \cdot \int e^{u^{\frac{4}{3}}} u^{\frac{1}{3}} d u

Apply u-substitution

: \int \frac{3 e ^{v}}{4} d v

= - 9 \cdot \frac{1}{3} \cdot \int \frac{3 e ^{v}}{4} d v

Take the constant out:

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

= - 9 \cdot \frac{1}{3} \cdot \frac{3}{4} \cdot \int e^{v} d v

Use the common integral:

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

= - 9 \cdot \frac{1}{3} \cdot \frac{3}{4} e^{v}

Substitute back

= - 9 \cdot \frac{1}{3} \cdot \frac{3}{4} e^{(x^{3})^{\frac{4}{3}}}

Simplify

- 9 \cdot \frac{1}{3} \cdot \frac{3}{4} e^{(x^{3})^{\frac{4}{3}}} : - \frac{9}{4} e^{x^{4}}

= - \frac{9}{4} e^{x^{4}}

Add a constant to the solution

= - \frac{9}{4} e^{x^{4}} + C

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What is the integral of-9x^3e^{x^4} ?
The integral of-9x^3e^{x^4} is -9/4 e^{x^4}+C