What is the integral of 1/(sqrt(-x^2+4x+12)) ?

The integral of 1/(sqrt(-x^2+4x+12)) is arcsin(1/4 (x-2))+C

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integral of 1/(sqrt(-x^2+4x+12))

\int \frac{1}{- x ^{2} + 4 x + 12} d x

arcsin (\frac{1}{4} (x - 2)) + C

Solution steps

\int \frac{1}{- x ^{2} + 4 x + 12} d x

Complete the square

- x^{2} + 4 x + 12 : - (x - 2)^{2} + 16

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

Simplify

= arcsin (\frac{1}{4} (x - 2))

Add a constant to the solution

= arcsin (\frac{1}{4} (x - 2)) + C

\frac{x ^{2} + 5 x}{x + 2}

n = 1 \sum \infty \frac{8}{n ^{5}}

e^{- 2 x}

y = (x + x^{- 1})^{3}

\frac{d}{d x} ((x + 2)^{\frac{1}{2}})

What is the integral of 1/(sqrt(-x^2+4x+12)) ?
The integral of 1/(sqrt(-x^2+4x+12)) is arcsin(1/4 (x-2))+C