What is the integral of 1/(sqrt(3-x^2-2x)) ?

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

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integral of 1/(sqrt(3-x^2-2x))

\int \frac{1}{3 - x ^{2} - 2 x} d x

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

Solution steps

\int \frac{1}{3 - x ^{2} - 2 x} d x

Complete the square

3 - x^{2} - 2 x : - (x + 1)^{2} + 4

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

Apply u-substitution

: \int \frac{1}{- u ^{2} + 4} d u

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

Simplify

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

Add a constant to the solution

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

\int tan (x - 7) \frac{1}{x - 7} d x

\int e^{a x + 7} d x

\frac{\partial}{\partial y} (- x)

f (x) = x^{4} - 17 x^{2} + 16, at x = 1

sin^{2} (x), sin^{3} (x), [0, π]

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