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

The integral of (arctan(3x))/(1+9x^2) is 1/6 arctan^2(3x)+C

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

\int \frac{arctan ( 3 x )}{1 + 9 x ^{2}} d x

\frac{1}{6} arctan^{2} (3 x) + C

Solution steps

\int \frac{arctan ( 3 x )}{1 + 9 x ^{2}} d x

Apply Integral Substitution

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

Apply u-substitution

: \int v d v

= \frac{1}{3} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{1}{3} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{3} \cdot \frac{arctan ^{2} ( 3 x )}{2}

Simplify

\frac{1}{3} \cdot \frac{arctan ^{2} ( 3 x )}{2} : \frac{1}{6} arctan^{2} (3 x)

= \frac{1}{6} arctan^{2} (3 x)

Add a constant to the solution

= \frac{1}{6} arctan^{2} (3 x) + C

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What is the integral of (arctan(3x))/(1+9x^2) ?
The integral of (arctan(3x))/(1+9x^2) is 1/6 arctan^2(3x)+C