What is the integral of 9x^4e^{x^5} ?

The integral of 9x^4e^{x^5} is 9/5 e^{x^5}+C

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integral of 9x^4e^{x^5}

\int 9 x^{4} e^{x^{5}} d x

\frac{9}{5} e^{x^{5}} + C

Solution steps

\int 9 x^{4} e^{x^{5}} d x

Take the constant out:

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

= 9 \cdot \int x^{4} e^{x^{5}} d x

Apply u-substitution

: \int \frac{e ^{u^{\frac{5}{4}}} u ^{\frac{1}{4}}}{4} d u

= 9 \cdot \int \frac{e ^{u^{\frac{5}{4}}} u ^{\frac{1}{4}}}{4} d u

Take the constant out:

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

= 9 \cdot \frac{1}{4} \cdot \int e^{u^{\frac{5}{4}}} u^{\frac{1}{4}} d u

Apply u-substitution

: \int \frac{4 e ^{v}}{5} d v

= 9 \cdot \frac{1}{4} \cdot \int \frac{4 e ^{v}}{5} d v

Take the constant out:

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

= 9 \cdot \frac{1}{4} \cdot \frac{4}{5} \cdot \int e^{v} d v

Use the common integral:

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

= 9 \cdot \frac{1}{4} \cdot \frac{4}{5} e^{v}

Substitute back

= 9 \cdot \frac{1}{4} \cdot \frac{4}{5} e^{(x^{4})^{\frac{5}{4}}}

Simplify

9 \cdot \frac{1}{4} \cdot \frac{4}{5} e^{(x^{4})^{\frac{5}{4}}} : \frac{9}{5} e^{x^{5}}

= \frac{9}{5} e^{x^{5}}

Add a constant to the solution

= \frac{9}{5} e^{x^{5}} + C

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What is the integral of 9x^4e^{x^5} ?
The integral of 9x^4e^{x^5} is 9/5 e^{x^5}+C