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

x \to \infty lim (8 x^{2})

f (x) = 5 tan (x), at x = 1.57

\int (cos (t))^{2} d t

x \to 0 lim (\frac{sin ( 3 x )}{sin ( 6 x )})

\int 3 e^{x} - 9 cos (x) + C 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