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

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

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integral of 1/(w^2+9)

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

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

Solution steps

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

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{w}{3}

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

Add a constant to the solution

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

\frac{d}{d x} (5 - 4 x)

\frac{d}{d x} (x^{2} - 3 x + 2)

\int 2 cos (x) cos^{5} (sin (x)) d x

\frac{d}{d x} (5 \frac{x y}{2})

2 x \frac{d y}{d x} = cos^{2} (y)

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