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

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

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

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

\frac{1}{3} e^{x^{3} + 3 x} + C

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Substitute back

u = x^{3} + 3 x

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

Add a constant to the solution

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

\int_{0}^{1} \frac{1}{x ^{0.4}} d x

\frac{1}{2} (8 t + 3)^{- \frac{1}{2}}

\frac{d}{d x} (f (x) \cdot f (x) \cdot f (x))

y^{^{'}} - \frac{8}{x} y = y^{3} \frac{1}{x ^{12}}

y = x^{2}, x = y^{2}

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