What is the integral of ((x+1)^2)/(e^x) ?

The integral of ((x+1)^2)/(e^x) is -(x^2)/(e^x)-(4x)/(e^x)-5/(e^x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((x+1)^2)/(e^x)

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

- \frac{x ^{2}}{e ^{x}} - \frac{4 x}{e ^{x}} - \frac{5}{e ^{x}} + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

- \frac{x ^{2}}{e ^{x}} + 2 (- \frac{x}{e ^{x}} - \frac{1}{e ^{x}}) + 2 (- \frac{x}{e ^{x}} - \frac{1}{e ^{x}}) - \frac{1}{e ^{x}} : - \frac{x ^{2}}{e ^{x}} - \frac{4 x}{e ^{x}} - \frac{5}{e ^{x}}

= - \frac{x ^{2}}{e ^{x}} - \frac{4 x}{e ^{x}} - \frac{5}{e ^{x}}

Add a constant to the solution

= - \frac{x ^{2}}{e ^{x}} - \frac{4 x}{e ^{x}} - \frac{5}{e ^{x}} + C

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

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

\int (x + 5) e^{x + 3} d x

\int sin (x) \cdot sin (x) d x

\int sec (4 x + 4) tan (4 x + 4) d x

What is the integral of ((x+1)^2)/(e^x) ?
The integral of ((x+1)^2)/(e^x) is -(x^2)/(e^x)-(4x)/(e^x)-5/(e^x)+C