What is the integral of 1/(sqrt(-x^2+8x-12)) ?

The integral of 1/(sqrt(-x^2+8x-12)) is arcsin(1/2 (x-4))+C

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integral of 1/(sqrt(-x^2+8x-12))

\int \frac{1}{- x ^{2} + 8 x - 12} d x

arcsin (\frac{1}{2} (x - 4)) + C

Solution steps

\int \frac{1}{- x ^{2} + 8 x - 12} d x

Complete the square

- x^{2} + 8 x - 12 : - (x - 4)^{2} + 4

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

Simplify

= arcsin (\frac{1}{2} (x - 4))

Add a constant to the solution

= arcsin (\frac{1}{2} (x - 4)) + C

\frac{d}{d x} (- \frac{I n ( x ^{3} )}{x})

x^{5} + x^{2} y + 9 y^{2} = 13

y = 4 x - x^{2}, y = 3 - x

\frac{\partial}{\partial x} (cos (3 x))

\int (sec (x) - \frac{1}{sin ( x )} + cos (x)) d x

What is the integral of 1/(sqrt(-x^2+8x-12)) ?
The integral of 1/(sqrt(-x^2+8x-12)) is arcsin(1/2 (x-4))+C