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

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

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

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

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

Solution steps

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

Complete the square

3 - 2 x - x^{2} : - (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

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

\frac{\partial}{\partial x} (\frac{7 x y ^{2}}{x ^{2} + y ^{4}})

\int \frac{cos ( x )}{sin ^{2} ( x ) + 2 sin ( x )} d x

\int 2 e^{- \frac{x}{4}} d x

x \to 9 + lim (6)

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