What is the integral of 5x^2e^{x^3} ?

The integral of 5x^2e^{x^3} is 5/3 e^{x^3}+C

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integral of 5x^2e^{x^3}

\int 5 x^{2} e^{x^{3}} d x

\frac{5}{3} e^{x^{3}} + C

Solution steps

\int 5 x^{2} e^{x^{3}} d x

Take the constant out:

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

= 5 \cdot \int x^{2} e^{x^{3}} d x

Apply u-substitution

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

= 5 \cdot \int \frac{e ^{u}}{3} d u

Take the constant out:

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

= 5 \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= 5 \cdot \frac{1}{3} e^{u}

Substitute back

u = x^{3}

= 5 \cdot \frac{1}{3} e^{x^{3}}

Simplify

5 \cdot \frac{1}{3} e^{x^{3}} : \frac{5}{3} e^{x^{3}}

= \frac{5}{3} e^{x^{3}}

Add a constant to the solution

= \frac{5}{3} e^{x^{3}} + C

\frac{d}{d x} ((2 x - 3)^{5} (x^{2} + 4 x)^{6})

\int tan^{4} (\frac{x}{3}) d x

\int_{1}^{2} \frac{x}{x ^{2} - 1} d x

\frac{∣ x ∣}{5 - x ^{2}}

x^{2} y d x - (x^{3} + y^{3}) d y = 0

What is the integral of 5x^2e^{x^3} ?
The integral of 5x^2e^{x^3} is 5/3 e^{x^3}+C