What is the integral of 1/((2x+4)sqrt(2x-5)) ?

The integral of 1/((2x+4)sqrt(2x-5)) is 1/3 arctan((sqrt(2x-5))/3)+C

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integral of 1/((2x+4)sqrt(2x-5))

\int \frac{1}{( 2 x + 4 ) 2 x - 5} d x

\frac{1}{3} arctan (\frac{2 x - 5}{3}) + C

Solution steps

\int \frac{1}{( 2 x + 4 ) 2 x - 5} d x

Apply u-substitution

= \int \frac{1}{2 u u - 9} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u u - 9} d u

Apply u-substitution

= \frac{1}{2} \cdot \int \frac{2}{v ^{2} + 9} d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{v ^{2} + 9} d v

Apply Integral Substitution

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{3 ( w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} arctan (w)

Substitute back

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} arctan (\frac{2 x + 4 - 9}{3})

Simplify

\frac{1}{2} \cdot 2 \cdot \frac{1}{3} arctan (\frac{2 x + 4 - 9}{3}) : \frac{1}{3} arctan (\frac{2 x - 5}{3})

= \frac{1}{3} arctan (\frac{2 x - 5}{3})

Add a constant to the solution

= \frac{1}{3} arctan (\frac{2 x - 5}{3}) + C

What is the integral of 1/((2x+4)sqrt(2x-5)) ?
The integral of 1/((2x+4)sqrt(2x-5)) is 1/3 arctan((sqrt(2x-5))/3)+C