What is the integral of x^2sqrt(e^{x^3+1)} ?

The integral of x^2sqrt(e^{x^3+1)} is 2/3 e^{1/2 (x^3+1)}+C

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integral of x^2sqrt(e^{x^3+1)}

\int x^{2} e^{x^{3} + 1} d x

\frac{2}{3} e^{\frac{1}{2} (x^{3} + 1)} + C

Solution steps

\int x^{2} e^{x^{3} + 1} d x

Apply u-substitution

: \int \frac{e ^{\frac{u}{2}}}{3} d u

= \int \frac{e ^{\frac{u}{2}}}{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^{\frac{u}{2}} d u

Apply u-substitution

: \int e^{v} \cdot 2 d v

= \frac{1}{3} \cdot \int e^{v} \cdot 2 d v

Take the constant out:

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

= \frac{1}{3} \cdot 2 \cdot \int e^{v} d v

Use the common integral:

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

= \frac{1}{3} \cdot 2 e^{v}

Substitute back

= \frac{1}{3} \cdot 2 e^{\frac{x ^{3} + 1}{2}}

Simplify

\frac{1}{3} \cdot 2 e^{\frac{x ^{3} + 1}{2}} : \frac{2}{3} e^{\frac{1}{2} (x^{3} + 1)}

= \frac{2}{3} e^{\frac{1}{2} (x^{3} + 1)}

Add a constant to the solution

= \frac{2}{3} e^{\frac{1}{2} (x^{3} + 1)} + C

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What is the integral of x^2sqrt(e^{x^3+1)} ?
The integral of x^2sqrt(e^{x^3+1)} is 2/3 e^{1/2 (x^3+1)}+C