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

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

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

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

\frac{2}{9} arctan^{\frac{3}{2}} (3 x) + C

Solution steps

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

Apply u-substitution

: \int \frac{2 u ^{2}}{3} d u

= \int \frac{2 u ^{2}}{3} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{2}{3} \cdot \frac{u ^{3}}{3}

Substitute back

u = arctan (3 x)

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

Simplify

\frac{2}{3} \cdot \frac{( arctan ( 3 x ) ) ^{3}}{3} : \frac{2}{9} arctan^{\frac{3}{2}} (3 x)

= \frac{2}{9} arctan^{\frac{3}{2}} (3 x)

Add a constant to the solution

= \frac{2}{9} arctan^{\frac{3}{2}} (3 x) + C

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