What is the integral of 6xarctan(3x^2) ?

The integral of 6xarctan(3x^2) is 6(1/2 x^2arctan(3x^2)-1/12 ln|9x^4+1|)+C

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integral of 6xarctan(3x^2)

\int 6 x arctan (3 x^{2}) d x

6 (\frac{1}{2} x^{2} arctan (3 x^{2}) - \frac{1}{12} ln 9 x^{4} + 1) + C

Solution steps

\int 6 x arctan (3 x^{2}) d x

Take the constant out:

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

= 6 \cdot \int x arctan (3 x^{2}) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} arctan (3 x^{2}) - \int \frac{3 x ^{3}}{9 x ^{4} + 1} d x

= 6 (\frac{1}{2} x^{2} arctan (3 x^{2}) - \int \frac{3 x ^{3}}{9 x ^{4} + 1} d x)

\int \frac{3 x ^{3}}{9 x ^{4} + 1} d x = \frac{1}{12} ln 9 x^{4} + 1

= 6 (\frac{1}{2} x^{2} arctan (3 x^{2}) - \frac{1}{12} ln 9 x^{4} + 1)

Add a constant to the solution

= 6 (\frac{1}{2} x^{2} arctan (3 x^{2}) - \frac{1}{12} ln 9 x^{4} + 1) + C

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What is the integral of 6xarctan(3x^2) ?
The integral of 6xarctan(3x^2) is 6(1/2 x^2arctan(3x^2)-1/12 ln|9x^4+1|)+C