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

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

Popular >

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

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

4 e^{x^{2} + x} + C

Solution steps

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

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

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

Take the constant out:

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

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

Apply u-substitution

: \int e^{u} d u

= 4 \cdot \int e^{u} d u

Use the common integral:

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

= 4 e^{u}

Substitute back

u = x^{2} + x

= 4 e^{x^{2} + x}

Add a constant to the solution

= 4 e^{x^{2} + x} + C

f (x) = x^{2} - tan (x)

\int_{0}^{1} \frac{x}{x ^{2} + 4 x + 4} d x

\frac{\partial}{\partial x} (e^{- 6 x^{2} - 3 y^{2}})

\frac{( x + 1 )}{x ^{2} + 1}

\int 3 x^{\frac{5}{2}} - x^{\frac{5}{3}} d x

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