What is the integral of 3/(sqrt(5-4u-u^2)) ?

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

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integral of 3/(sqrt(5-4u-u^2))

\int \frac{3}{5 - 4 u - u ^{2}} d u

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

Solution steps

\int \frac{3}{5 - 4 u - u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{1}{5 - 4 u - u ^{2}} d u

Complete the square

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

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

Apply u-substitution

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

= 3 \cdot \int \frac{1}{- v ^{2} + 9} d v

Apply Trigonometric Substitution

: \int 1 d w

= 3 \cdot \int 1 d w

Integral of a constant:

\int a d x = a x

= 3 \cdot 1 \cdot w

Substitute back

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

Simplify

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

Add a constant to the solution

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

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\int_{0}^{1} \frac{9}{x ( 1 + x )} d x

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