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

The integral of (sqrt(2x-3))/(2x+1) is -2arctan(1/2 sqrt(2x-3))+sqrt(2x-3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int \frac{2 x - 3}{2 x + 1} d x

- 2 arctan (\frac{1}{2} 2 x - 3) + 2 x - 3 + C

Solution steps

\int \frac{2 x - 3}{2 x + 1} d x

Apply u-substitution

: \int \frac{u ^{2}}{u ^{2} + 4} d u

= \int \frac{u ^{2}}{u ^{2} + 4} d u

\frac{u ^{2}}{u ^{2} + 4} = - \frac{4}{u ^{2} + 4} + 1

= \int - \frac{4}{u ^{2} + 4} + 1 d u

Apply Integral Substitution

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

Take the constant out:

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

= 2 \cdot \int - \frac{1}{v ^{2} + 1} + 1 d v

Apply the Sum Rule:

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

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

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

\int 1 d v = v

= 2 (- arctan (v) + v)

Substitute back

= 2 (- arctan (\frac{2 x - 3}{2}) + \frac{2 x - 3}{2})

Simplify

2 (- arctan (\frac{2 x - 3}{2}) + \frac{2 x - 3}{2}) : - 2 arctan (\frac{1}{2} 2 x - 3) + 2 x - 3

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

Add a constant to the solution

= - 2 arctan (\frac{1}{2} 2 x - 3) + 2 x - 3 + C

\int - 4 sin (5 x) sin (x) d x

\frac{d}{d x} (\frac{1}{1 + sin ( x )})

\int \frac{3}{4} t + 6 d t

x \to 3 - lim (9 - x^{2})

f (x) = \frac{x + 2}{x ^{2} - 9}, at x = 1

What is the integral of (sqrt(2x-3))/(2x+1) ?
The integral of (sqrt(2x-3))/(2x+1) is -2arctan(1/2 sqrt(2x-3))+sqrt(2x-3)+C