What is the integral of 1/(sqrt(3+4x-4x^2)) ?

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

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integral of 1/(sqrt(3+4x-4x^2))

\int \frac{1}{3 + 4 x - 4 x ^{2}} d x

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

Solution steps

\int \frac{1}{3 + 4 x - 4 x ^{2}} d x

Complete the square

3 + 4 x - 4 x^{2} : - 4 (x - \frac{1}{2})^{2} + 4

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = x - \frac{1}{2}

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

Add a constant to the solution

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

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