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

The integral of 1/(sqrt(5-x^2-4x)) is arcsin(1/3 (x+2))+C

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integral of 1/(sqrt(5-x^2-4x))

\int \frac{1}{5 - x ^{2} - 4 x} d x

arcsin (\frac{1}{3} (x + 2)) + C

Solution steps

\int \frac{1}{5 - x ^{2} - 4 x} d x

Complete the square

5 - x^{2} - 4 x : - (x + 2)^{2} + 9

= \int \frac{1}{- ( x + 2 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 9} d u

= \int \frac{1}{- u ^{2} + 9} 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}{3} (x + 2))

Simplify

= arcsin (\frac{1}{3} (x + 2))

Add a constant to the solution

= arcsin (\frac{1}{3} (x + 2)) + C

x^{3} \frac{d y}{d x} = y^{2} + 1

\int \frac{12}{( x ^{2} - 6 x + 13 ) ^{2}} d x

\int (x^{2} - 9)^{3} (2 x) d x

y = 8 x, (16, 32)

\frac{\partial}{\partial x} (x + \frac{1}{x})

What is the integral of 1/(sqrt(5-x^2-4x)) ?
The integral of 1/(sqrt(5-x^2-4x)) is arcsin(1/3 (x+2))+C