What is the integral of (2x+3)sqrt(2x-1) ?

The integral of (2x+3)sqrt(2x-1) is 2/5 x(2x-1)^{3/2}+17/15 (2x-1)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x+3)sqrt(2x-1)

\int (2 x + 3) 2 x - 1 d x

\frac{2}{5} x (2 x - 1)^{\frac{3}{2}} + \frac{17}{15} (2 x - 1)^{\frac{3}{2}} + C

Solution steps

\int (2 x + 3) 2 x - 1 d x

Expand

(2 x + 3) 2 x - 1 : 2 x 2 x - 1 + 3 2 x - 1

= \int 2 x 2 x - 1 + 3 2 x - 1 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 2 x 2 x - 1 d x + \int 3 2 x - 1 d x

\int 2 x 2 x - 1 d x = \frac{2}{15} (3 x + 1) (2 x - 1)^{\frac{3}{2}}

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

= \frac{2}{15} (3 x + 1) (2 x - 1)^{\frac{3}{2}} + (2 x - 1)^{\frac{3}{2}}

Simplify

\frac{2}{15} (3 x + 1) (2 x - 1)^{\frac{3}{2}} + (2 x - 1)^{\frac{3}{2}} : \frac{2}{5} x (2 x - 1)^{\frac{3}{2}} + \frac{17}{15} (2 x - 1)^{\frac{3}{2}}

= \frac{2}{5} x (2 x - 1)^{\frac{3}{2}} + \frac{17}{15} (2 x - 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} x (2 x - 1)^{\frac{3}{2}} + \frac{17}{15} (2 x - 1)^{\frac{3}{2}} + C

\int 3 x^{3} - \frac{1}{x} + 5 e^{x} d x

\int \frac{1}{k x} d x

\int \frac{x - 4}{x ^{2} + x - 2} d x

\int x^{2} (5 x^{3} - 3)^{7} d x

\int \frac{6 x + 4}{x ^{2} + 1} d x

What is the integral of (2x+3)sqrt(2x-1) ?
The integral of (2x+3)sqrt(2x-1) is 2/5 x(2x-1)^{3/2}+17/15 (2x-1)^{3/2}+C