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

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

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

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

2 x - 3 + 1 - ln 2 x - 3 + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{u}{u + 1} d u

= \int \frac{u}{u + 1} d u

Apply u-substitution

: \int \frac{v - 1}{v} d v

= \int \frac{v - 1}{v} d v

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= \int 1 - \frac{1}{v} 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

= \int 1 d v - \int \frac{1}{v} d v

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= v - ln ∣ v ∣

Substitute back

= 2 x - 3 + 1 - ln 2 x - 3 + 1

Add a constant to the solution

= 2 x - 3 + 1 - ln 2 x - 3 + 1 + C

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