What is the integral of xarctan(4x) ?

The integral of xarctan(4x) is 1/2 x^2arctan(4x)-1/32 (-arctan(4x)+4x)+C

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integral of xarctan(4x)

\int x arctan (4 x) d x

\frac{1}{2} x^{2} arctan (4 x) - \frac{1}{32} (- arctan (4 x) + 4 x) + C

Solution steps

\int x arctan (4 x) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} arctan (4 x) - \int \frac{2 x ^{2}}{16 x ^{2} + 1} d x

= \frac{1}{2} x^{2} arctan (4 x) - \int \frac{2 x ^{2}}{16 x ^{2} + 1} d x

\int \frac{2 x ^{2}}{16 x ^{2} + 1} d x = \frac{1}{32} (- arctan (4 x) + 4 x)

= \frac{1}{2} x^{2} arctan (4 x) - \frac{1}{32} (- arctan (4 x) + 4 x)

Add a constant to the solution

= \frac{1}{2} x^{2} arctan (4 x) - \frac{1}{32} (- arctan (4 x) + 4 x) + C

x^{2} - 4 x + 1

\int sin (x) cos^{6} (x) d x

x \to 0 lim (cos^{2} (x))

\frac{x ^{2} + 1}{x - 2}

\frac{d}{d x} ((x^{4} - 2)^{5})

What is the integral of xarctan(4x) ?
The integral of xarctan(4x) is 1/2 x^2arctan(4x)-1/32 (-arctan(4x)+4x)+C