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

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

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

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

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

Solution steps

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

Complete the square

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

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

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} (r + 2))

Simplify

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

Add a constant to the solution

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

\int e^{x} - e^{a} (x - a) d x

\int \frac{3 x ^{3} + 8 x}{( x ^{2} + 2 ) ^{2}} d x

\frac{\partial}{\partial x} (p (q, x) q)

\frac{d}{d x} (x (2 x + 3)^{4})

\int \frac{1}{x} x^{'} d x

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