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

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

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

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

arcsin (x - 2) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Use the common integral:

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

= arcsin (u)

Substitute back

u = x - 2

= arcsin (x - 2)

Add a constant to the solution

= arcsin (x - 2) + C

\int \frac{v + 1}{2 v + 1} d v

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\int \frac{x ^{2} + x + 8}{x ^{2} + 3 x - 4} d x

\int - \frac{2}{49 - 9 x ^{2}} d x

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