What is the integral of 3/(x^2+81) ?

The integral of 3/(x^2+81) is 1/3 arctan(x/9)+C

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integral of 3/(x^2+81)

\int \frac{3}{x ^{2} + 81} d x

\frac{1}{3} arctan (\frac{x}{9}) + C

Solution steps

\int \frac{3}{x ^{2} + 81} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{x ^{2} + 81} d x

Apply Integral Substitution

= 3 \cdot \int \frac{1}{9 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 3 \cdot \frac{1}{9} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 3 \cdot \frac{1}{9} arctan (u)

Substitute back

u = \frac{x}{9}

= 3 \cdot \frac{1}{9} arctan (\frac{x}{9})

Simplify

3 \cdot \frac{1}{9} arctan (\frac{x}{9}) : \frac{1}{3} arctan (\frac{x}{9})

= \frac{1}{3} arctan (\frac{x}{9})

Add a constant to the solution

= \frac{1}{3} arctan (\frac{x}{9}) + C

\frac{cos ( x )}{x ^{2}}

\int \frac{\frac{1}{x}}{x ^{2} + 4} d x

\int \frac{1}{3} x e^{\frac{x}{6}} d x

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\int \frac{10}{x ^{2} x ^{2} + 25} d x

What is the integral of 3/(x^2+81) ?
The integral of 3/(x^2+81) is 1/3 arctan(x/9)+C