What is the integral of arctan(3x-1) ?

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

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integral of arctan(3x-1)

\int arctan (3 x - 1) d x

\frac{1}{6} (2 arctan (3 x - 1) (3 x - 1) - ln 9 x^{2} - 6 x + 2) + C

Solution steps

\int arctan (3 x - 1) d x

Apply u-substitution

: \int arctan (u) \frac{1}{3} d u

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

Apply Integration By Parts

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

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

\int \frac{u}{u ^{2} + 1} d u = \frac{1}{2} ln u^{2} + 1

= \frac{1}{3} (u arctan (u) - \frac{1}{2} ln u^{2} + 1)

Substitute back

u = 3 x - 1

= \frac{1}{3} ((3 x - 1) arctan (3 x - 1) - \frac{1}{2} ln (3 x - 1)^{2} + 1)

Simplify

\frac{1}{3} ((3 x - 1) arctan (3 x - 1) - \frac{1}{2} ln (3 x - 1)^{2} + 1) : \frac{1}{6} (2 arctan (3 x - 1) (3 x - 1) - ln 9 x^{2} - 6 x + 2)

= \frac{1}{6} (2 arctan (3 x - 1) (3 x - 1) - ln 9 x^{2} - 6 x + 2)

Add a constant to the solution

= \frac{1}{6} (2 arctan (3 x - 1) (3 x - 1) - ln 9 x^{2} - 6 x + 2) + C

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