What is the integral of 1/(sqrt(-x^2+6x-5)) ?

The integral of 1/(sqrt(-x^2+6x-5)) is arcsin(1/2 (x-3))+C

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integral of 1/(sqrt(-x^2+6x-5))

\int \frac{1}{- x ^{2} + 6 x - 5} d x

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

Solution steps

\int \frac{1}{- x ^{2} + 6 x - 5} d x

Complete the square

- x^{2} + 6 x - 5 : - (x - 3)^{2} + 4

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

Apply u-substitution

: \int \frac{1}{- u ^{2} + 4} d u

= \int \frac{1}{- u ^{2} + 4} d u

Apply Trigonometric Substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

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

Simplify

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

Add a constant to the solution

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

4 x^{3} y + x^{4} y^{^{'}} = sin^{3} (\frac{x}{9})

ln (x + 1), 0

\frac{d}{d x} (3 sin (2 x) cos (x))

\frac{\partial}{\partial t} (1)

f (x) = \frac{x ^{2} + 2 x + 3}{x ^{2} + 2 x + 1}

What is the integral of 1/(sqrt(-x^2+6x-5)) ?
The integral of 1/(sqrt(-x^2+6x-5)) is arcsin(1/2 (x-3))+C