What is the integral of 1/(sqrt(-x^2+2x+15)) ?

The integral of 1/(sqrt(-x^2+2x+15)) is arcsin(1/4 (x-1))+C

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integral of 1/(sqrt(-x^2+2x+15))

\int \frac{1}{- x ^{2} + 2 x + 15} d x

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

Solution steps

\int \frac{1}{- x ^{2} + 2 x + 15} d x

Complete the square

- x^{2} + 2 x + 15 : - (x - 1)^{2} + 16

= \int \frac{1}{- ( x - 1 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 16} d u

= \int \frac{1}{- u ^{2} + 16} 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}{4} (x - 1))

Simplify

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

Add a constant to the solution

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

\frac{6}{s} + \frac{32}{s ^{2} + 16}

\frac{8}{( s ^{2} + 1 ) ^{2}}

x^{2} \frac{d y}{d x} - y^{2} = 0

x \to 0 lim (x (ln (x))^{2})

\int \frac{4}{2 x + 1} d x

What is the integral of 1/(sqrt(-x^2+2x+15)) ?
The integral of 1/(sqrt(-x^2+2x+15)) is arcsin(1/4 (x-1))+C