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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{4 ( - x ^{2} - 4 x - 4 )}{e ^{x + 1}} + C

Solution steps

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

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

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

Take the constant out:

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

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

Simplify

= \frac{4}{e} \cdot \int \frac{x ^{2} + 2 x}{e ^{x}} d x

Expand

\frac{x ^{2} + 2 x}{e ^{x}} : \frac{x ^{2}}{e ^{x}} + \frac{2 x}{e ^{x}}

= \frac{4}{e} \cdot \int \frac{x ^{2}}{e ^{x}} + \frac{2 x}{e ^{x}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{2 x}{e ^{x}} d x = 2 (- \frac{x}{e ^{x}} - \frac{1}{e ^{x}})

= \frac{4}{e} (- \frac{x ^{2}}{e ^{x}} + 2 (- \frac{x}{e ^{x}} - \frac{1}{e ^{x}}) + 2 (- \frac{x}{e ^{x}} - \frac{1}{e ^{x}}))

Simplify

\frac{4}{e} (- \frac{x ^{2}}{e ^{x}} + 2 (- \frac{x}{e ^{x}} - \frac{1}{e ^{x}}) + 2 (- \frac{x}{e ^{x}} - \frac{1}{e ^{x}})) : \frac{4 ( - x ^{2} - 4 x - 4 )}{e ^{x + 1}}

= \frac{4 ( - x ^{2} - 4 x - 4 )}{e ^{x + 1}}

Add a constant to the solution

= \frac{4 ( - x ^{2} - 4 x - 4 )}{e ^{x + 1}} + C

\int \frac{z ^{5}}{( 9 z ^{2} - 25 ) ^{\frac{3}{2}}} d z

\int \frac{x}{x ^{'}} d x

\int \frac{4 + x ^{6}}{x} d x

\int \frac{1 - sin ^{2} ( x )}{3 cos ^{2} ( x )} d x

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

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