What is the integral of 1/(sqrt(3-4x-x^2)) ?

The integral of 1/(sqrt(3-4x-x^2)) is arcsin(1/(sqrt(7))(x+2))+C

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integral of 1/(sqrt(3-4x-x^2))

\int \frac{1}{3 - 4 x - x ^{2}} d x

arcsin (\frac{1}{7} (x + 2)) + C

Solution steps

\int \frac{1}{3 - 4 x - x ^{2}} d x

Complete the square

3 - 4 x - x^{2} : - (x + 2)^{2} + 7

= \int \frac{1}{- ( x + 2 ) ^{2} + 7} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 7} d u

= \int \frac{1}{- u ^{2} + 7} 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}{7} (x + 2))

Simplify

= arcsin (\frac{1}{7} (x + 2))

Add a constant to the solution

= arcsin (\frac{1}{7} (x + 2)) + C

d y = (x - 2 y cot (2 x)) d x

x (8 x + 5)^{4}

x^{3} e^{x} cos (x)

\int \frac{2 x + 4}{x ^{2} - 2 x - 3} d x

\int 3 x^{3} + 4 x^{2} - 2 x^{- 1} + 13 d x

What is the integral of 1/(sqrt(3-4x-x^2)) ?
The integral of 1/(sqrt(3-4x-x^2)) is arcsin(1/(sqrt(7))(x+2))+C