What is the integral of (2u)/(sqrt(-4u^2+9)) ?

The integral of (2u)/(sqrt(-4u^2+9)) is -1/2 sqrt(-4u^2+9)+C

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integral of (2u)/(sqrt(-4u^2+9))

\int \frac{2 u}{- 4 u ^{2} + 9} d u

- \frac{1}{2} - 4 u^{2} + 9 + C

Solution steps

\int \frac{2 u}{- 4 u ^{2} + 9} d u

Take the constant out:

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

= 2 \cdot \int \frac{u}{- 4 u ^{2} + 9} d u

Apply u-substitution

: \int - \frac{1}{8 v} d v

= 2 \cdot \int - \frac{1}{8 v} d v

Take the constant out:

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

= 2 (- \frac{1}{8} \cdot \int \frac{1}{v} d v)

Apply the Power Rule

: 2 v^{\frac{1}{2}}

= 2 (- \frac{1}{8} \cdot 2 v^{\frac{1}{2}})

Substitute back

v = - 4 u^{2} + 9

= 2 (- \frac{1}{8} \cdot 2 (- 4 u^{2} + 9)^{\frac{1}{2}})

Simplify

2 (- \frac{1}{8} \cdot 2 (- 4 u^{2} + 9)^{\frac{1}{2}}) : - \frac{1}{2} - 4 u^{2} + 9

= - \frac{1}{2} - 4 u^{2} + 9

Add a constant to the solution

= - \frac{1}{2} - 4 u^{2} + 9 + C

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What is the integral of (2u)/(sqrt(-4u^2+9)) ?
The integral of (2u)/(sqrt(-4u^2+9)) is -1/2 sqrt(-4u^2+9)+C