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

The integral of 1/(4u^2+9) is 1/6 arctan(2/3 u)+C

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integral of 1/(4u^2+9)

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

\frac{1}{6} arctan (\frac{2}{3} u) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{6 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{6} arctan (v)

Substitute back

v = \frac{2}{3} u

= \frac{1}{6} arctan (\frac{2}{3} u)

Add a constant to the solution

= \frac{1}{6} arctan (\frac{2}{3} u) + C

What is the integral of 1/(4u^2+9) ?
The integral of 1/(4u^2+9) is 1/6 arctan(2/3 u)+C