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

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

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

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

- - u^{2} + 9 + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2 v} d v

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

Take the constant out:

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

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

Apply the Power Rule

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

= - \frac{1}{2} \cdot 2 v^{\frac{1}{2}}

Substitute back

v = - u^{2} + 9

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

Simplify

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

= - - u^{2} + 9

Add a constant to the solution

= - - u^{2} + 9 + C

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