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

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

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

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

arcsin (x - 2) + C

Solution steps

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

Complete the square

- x^{2} + 4 x - 3 : - (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{15}{4} x^{2} d x

\frac{d}{d t} (lo g_{3} (2 t - x))

y = 2 x^{3} + 3 x^{2} - 12 x + 1

\frac{d}{d r} (r + 3 r)

\frac{\partial}{\partial y} (\frac{2 y}{x ^{2} - y ^{2}})

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