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

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

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

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

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

Solution steps

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

Complete the square

15 + 2 x - x^{2} : - (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{d}{d x} (cos (arcsin (x)))

\frac{\partial}{\partial x} (tan (x))

6 x^{2}, 0, 1, 5

\int_{- 1}^{3} (2 x)^{3} d x

(7) (0 - 6)

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