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

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

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

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

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

Solution steps

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

Complete the square

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

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

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} (r + 2))

Simplify

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

Add a constant to the solution

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

f (x) = \frac{x}{1 + 5 x ^{2}}, at x = 3

y = x + 3

\int 5 sin^{3} (6 x) cos^{3} (6 x) d x

\int \frac{5 x ^{5}}{5 + x ^{6}} d x

4 x^{2} + 2 x, at - 3

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