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

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

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integral of u 1/(sqrt(4-9u^2))

\int u \frac{1}{4 - 9 u ^{2}} d u

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

Solution steps

\int u \frac{1}{4 - 9 u ^{2}} d u

Simplify

u \frac{1}{4 - 9 u ^{2}} : \frac{u}{4 - 9 u ^{2}}

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

Apply u-substitution

: \int - \frac{1}{18 v} d v

= \int - \frac{1}{18 v} d v

Take the constant out:

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

= - \frac{1}{18} \cdot \int \frac{1}{v} d v

Apply the Power Rule

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

= - \frac{1}{18} \cdot 2 v^{\frac{1}{2}}

Substitute back

v = 4 - 9 u^{2}

= - \frac{1}{18} \cdot 2 (4 - 9 u^{2})^{\frac{1}{2}}

Simplify

- \frac{1}{18} \cdot 2 (4 - 9 u^{2})^{\frac{1}{2}} : - \frac{1}{9} 4 - 9 u^{2}

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

Add a constant to the solution

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

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