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

The integral of 1/(sqrt(-x^2+2x+8)) is arcsin(1/3 (x-1))+C

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

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

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

Solution steps

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

Complete the square

- x^{2} + 2 x + 8 : - (x - 1)^{2} + 9

= \int \frac{1}{- ( x - 1 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 9} d u

= \int \frac{1}{- u ^{2} + 9} 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}{3} (x - 1))

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (2 x y + y cos (x))

e^{8 x} - e^{2 x}, x = 1

y = x - 6, y^{2} = 2 x + 108

x \to 3 lim (\frac{13 sin ( 6 x )}{x})

What is the integral of 1/(sqrt(-x^2+2x+8)) ?
The integral of 1/(sqrt(-x^2+2x+8)) is arcsin(1/3 (x-1))+C