What is the integral of (24x+24)e^{4x^2+8x+b} ?

The integral of (24x+24)e^{4x^2+8x+b} is 3e^{b+4x^2+8x}+C

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integral of (24x+24)e^{4x^2+8x+b}

\int (24 x + 24) e^{4 x^{2} + 8 x + b} d x

3 e^{b + 4 x^{2} + 8 x} + C

Solution steps

\int (24 x + 24) e^{4 x^{2} + 8 x + b} d x

(24 x + 24) e^{4 x^{2} + 8 x + b} = 24 e^{b} (x + 1) e^{4 x^{2} + 8 x}

= \int 24 e^{b} (x + 1) e^{4 x^{2} + 8 x} d x

Take the constant out:

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

= 24 e^{b} \cdot \int (x + 1) e^{4 x^{2} + 8 x} d x

Apply u-substitution

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

= 24 e^{b} \cdot \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

= 24 e^{b} \frac{1}{8} \cdot \int e^{u} d u

Use the common integral:

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

= 24 e^{b} \frac{1}{8} e^{u}

Substitute back

u = 4 x^{2} + 8 x

= 24 e^{b} \frac{1}{8} e^{4 x^{2} + 8 x}

Simplify

24 e^{b} \frac{1}{8} e^{4 x^{2} + 8 x} : 3 e^{b + 4 x^{2} + 8 x}

= 3 e^{b + 4 x^{2} + 8 x}

Add a constant to the solution

= 3 e^{b + 4 x^{2} + 8 x} + C

\frac{d}{d x} ((x^{3} + 4) e^{3 x})

\int e^{x + y^{2} - z} d x

\int \frac{ln ( - 2 x )}{5 x} d x

t \to 0 lim (\frac{e ^{t} - 1}{t ^{9}})

y = x^{2} - 7, y = 6 x

What is the integral of (24x+24)e^{4x^2+8x+b} ?
The integral of (24x+24)e^{4x^2+8x+b} is 3e^{b+4x^2+8x}+C