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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 1 - u ^{2}} d u

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

Use the common integral:

\int \frac{1}{1 - u ^{2}} d u = arcsin (u)

= \frac{1}{2} arcsin (u)

Substitute back

u = 2 x - 3

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

Add a constant to the solution

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

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