What is the integral of (7x^2)e^{x^3+1} ?

The integral of (7x^2)e^{x^3+1} is 7/3 e^{x^3+1}+C

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integral of (7x^2)e^{x^3+1}

\int (7 x^{2}) e^{x^{3} + 1} d x

\frac{7}{3} e^{x^{3} + 1} + C

Solution steps

\int 7 x^{2} e^{x^{3} + 1} d x

Take the constant out:

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

= 7 \cdot \int x^{2} e^{x^{3} + 1} d x

Apply u-substitution

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

= 7 \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

= 7 \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= 7 \cdot \frac{1}{3} e^{u}

Substitute back

u = x^{3} + 1

= 7 \cdot \frac{1}{3} e^{x^{3} + 1}

Simplify

7 \cdot \frac{1}{3} e^{x^{3} + 1} : \frac{7}{3} e^{x^{3} + 1}

= \frac{7}{3} e^{x^{3} + 1}

Add a constant to the solution

= \frac{7}{3} e^{x^{3} + 1} + C

\int e^{- 2 y} y d y

f (x) = - x^{3} + x^{2} + 4 x - 2, at x = 0

\frac{\partial}{\partial y} (y e^{x y} sin (y))

x \to 2 - lim (3 x)

x \to 0 + lim (x^{3} cot (x))

What is the integral of (7x^2)e^{x^3+1} ?
The integral of (7x^2)e^{x^3+1} is 7/3 e^{x^3+1}+C