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

The integral of 1/(t^2+9) is 1/3 arctan(t/3)+C

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integral of 1/(t^2+9)

\int \frac{1}{t ^{2} + 9} d t

\frac{1}{3} arctan (\frac{t}{3}) + C

Solution steps

\int \frac{1}{t ^{2} + 9} d t

Apply Integral Substitution

= \int \frac{1}{3 ( 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}{3} \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}{3} arctan (u)

Substitute back

u = \frac{t}{3}

= \frac{1}{3} arctan (\frac{t}{3})

Add a constant to the solution

= \frac{1}{3} arctan (\frac{t}{3}) + C

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