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

The integral of 1/(sqrt(1-4x-x^2)) is arcsin(1/(sqrt(5))(x+2))+C

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

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

arcsin (\frac{1}{5} (x + 2)) + C

Solution steps

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

Complete the square

1 - 4 x - x^{2} : - (x + 2)^{2} + 5

= \int \frac{1}{- ( x + 2 ) ^{2} + 5} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 5} d u

= \int \frac{1}{- u ^{2} + 5} 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}{5} (x + 2))

Simplify

= arcsin (\frac{1}{5} (x + 2))

Add a constant to the solution

= arcsin (\frac{1}{5} (x + 2)) + C

\frac{d}{d x} (x^{\frac{7}{8}})

\int (csc^{2} (x) - a^{x}) d x

x \to \infty lim (0^{- x})

y^{^{''}} - 2 y^{^{'}} + 1 y = 0, y (0) = 5, y^{^{'}} (0) = 4

ln (x - 9)

What is the integral of 1/(sqrt(1-4x-x^2)) ?
The integral of 1/(sqrt(1-4x-x^2)) is arcsin(1/(sqrt(5))(x+2))+C