What is the integral of (17xe^{2x})/((1+2x)^2) ?

The integral of (17xe^{2x})/((1+2x)^2) is 17(-(e^{2x}x)/(2(1+2x))+1/4 e^{2x})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (17xe^{2x})/((1+2x)^2)

\int \frac{17 x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

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

Solution steps

\int \frac{17 x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

Take the constant out:

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

= 17 \cdot \int \frac{x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

Apply Integration By Parts

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

= 17 (- \frac{e ^{2 x} x}{2 ( 1 + 2 x )} - \int - \frac{1}{2} e^{2 x} d x)

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

= 17 (- \frac{e ^{2 x} x}{2 ( 1 + 2 x )} - (- \frac{1}{4} e^{2 x}))

Simplify

= 17 (- \frac{e ^{2 x} x}{2 ( 1 + 2 x )} + \frac{1}{4} e^{2 x})

Add a constant to the solution

= 17 (- \frac{e ^{2 x} x}{2 ( 1 + 2 x )} + \frac{1}{4} e^{2 x}) + C

\int cos^{6} (θ) sin (θ) d θ

\frac{\partial}{\partial y} (y^{3} sin (2 x))

\frac{d y}{d x} = 4

\frac{d}{d x} (4 x^{\frac{3}{5}})

\int_{1}^{2} (- 12 x + 6) d x

What is the integral of (17xe^{2x})/((1+2x)^2) ?
The integral of (17xe^{2x})/((1+2x)^2) is 17(-(e^{2x}x)/(2(1+2x))+1/4 e^{2x})+C