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

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

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

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

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

Solution steps

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

Simplify

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

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

Add a constant to the solution

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

x \to 0 - lim (x e^{\frac{1}{x}})

\int \frac{1}{x ( 1 - x )} d x

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

\frac{d y}{d x} = 3 x^{2} y^{2}

x \to 1 - lim (\frac{2}{x ^{3} - 1})

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