What is the integral of 1/(9x^2+16) ?

The integral of 1/(9x^2+16) is 1/12 arctan(3/4 x)+C

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integral of 1/(9x^2+16)

\int \frac{1}{9 x ^{2} + 16} d x

\frac{1}{12} arctan (\frac{3}{4} x) + C

Solution steps

\int \frac{1}{9 x ^{2} + 16} d x

Apply Integral Substitution

= \int \frac{1}{12 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{12} arctan (u)

Substitute back

u = \frac{3}{4} x

= \frac{1}{12} arctan (\frac{3}{4} x)

Add a constant to the solution

= \frac{1}{12} arctan (\frac{3}{4} x) + C

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What is the integral of 1/(9x^2+16) ?
The integral of 1/(9x^2+16) is 1/12 arctan(3/4 x)+C