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

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

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

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

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

Solution steps

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

Apply Integral Substitution

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

Use the common integral:

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

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

Substitute back

u = 3 x

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

Add a constant to the solution

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

9 \cdot t + 1

\frac{d}{d x} (x e^{- \frac{x}{4}})

\frac{d}{d x} (\frac{y}{y - x})

\frac{d}{d x} (e^{5 x + 1} arccsch (x + 1))

n = 1 \sum \infty \frac{2}{3 ^{2 n}}

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