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

The integral of 4/(sqrt(-u^2+9)) is 4arcsin(1/3 u)+C

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

= 4 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 4 \cdot 1 \cdot v

Substitute back

v = arcsin (\frac{1}{3} u)

= 4 \cdot 1 \cdot arcsin (\frac{1}{3} u)

Simplify

= 4 arcsin (\frac{1}{3} u)

Add a constant to the solution

= 4 arcsin (\frac{1}{3} u) + C

What is the integral of 4/(sqrt(-u^2+9)) ?
The integral of 4/(sqrt(-u^2+9)) is 4arcsin(1/3 u)+C