What is the integral of (x+1)e^{8x^2+16x} ?

The integral of (x+1)e^{8x^2+16x} is 1/16 e^{8x^2+16x}+C

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integral of (x+1)e^{8x^2+16x}

\int (x + 1) e^{8 x^{2} + 16 x} d x

\frac{1}{16} e^{8 x^{2} + 16 x} + C

Solution steps

\int (x + 1) e^{8 x^{2} + 16 x} d x

Apply u-substitution

: \int \frac{e ^{u}}{16} d u

= \int \frac{e ^{u}}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{16} e^{u}

Substitute back

u = 8 x^{2} + 16 x

= \frac{1}{16} e^{8 x^{2} + 16 x}

Add a constant to the solution

= \frac{1}{16} e^{8 x^{2} + 16 x} + C

\frac{\partial}{\partial x} (\frac{1}{2 x + y})

x \to 5 lim (\frac{x - 5}{x ^{2} - 5})

\int \frac{tan ^{4} ( x )}{cos ^{2} ( x )} d x

\frac{d y}{d x} = y e^{x}, y (0) = 2 e

\frac{d}{d x} (sin (y x))

What is the integral of (x+1)e^{8x^2+16x} ?
The integral of (x+1)e^{8x^2+16x} is 1/16 e^{8x^2+16x}+C