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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

\frac{d}{d x} (ln (e^{3 x} + e^{- 3 x}))

ln (\frac{x + 3}{x - 1})

\frac{d}{d x} (5 sec (4 x))

\int (3 x^{2} + y^{2}) ln (x^{2} + y^{2}) d x

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

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