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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Expand

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

= \int e^{2 x} + e^{x} x + e^{x} + 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 e^{2 x} d x + \int e^{x} x d x + \int e^{x} d x + \int x d x

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

x^{4} (2 x - 5)^{6}

y^{^{''}} + 40 y^{^{'}} + 625 y = 0

\frac{s ^{2} + 1}{s ^{2} + 4}

\int_{0}^{1} \frac{17}{x ^{5}} d x

\int x x^{2} + 2 x + 2 d x

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