What is the integral of (x+1)e^{4x^2+8x} ?

The integral of (x+1)e^{4x^2+8x} is 1/8 e^{4x^2+8x}+C

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integral of (x+1)e^{4x^2+8x}

\int (x + 1) e^{4 x^{2} + 8 x} d x

\frac{1}{8} e^{4 x^{2} + 8 x} + C

Solution steps

\int (x + 1) e^{4 x^{2} + 8 x} d x

Apply u-substitution

: \int \frac{e ^{u}}{8} d u

= \int \frac{e ^{u}}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{8} e^{u}

Substitute back

u = 4 x^{2} + 8 x

= \frac{1}{8} e^{4 x^{2} + 8 x}

Add a constant to the solution

= \frac{1}{8} e^{4 x^{2} + 8 x} + C

\int [\frac{( x ^{2} + 10 x + 25 )}{( x + 5 )}] d x

\int \frac{e ^{x}}{1 + 81 e ^{2 x}} d x

f (x) = 1 + sin (x)

\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x y}

9 x^{2}

What is the integral of (x+1)e^{4x^2+8x} ?
The integral of (x+1)e^{4x^2+8x} is 1/8 e^{4x^2+8x}+C