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

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

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

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

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

Solution steps

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

Complete the square

8 - 2 x - x^{2} : - (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{d}{d x} (\frac{3}{2})

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

\int e^{- 3 x} sin (b x) d x

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

\int e^{4 x - 9} d x

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