What is the integral of 4xarctan(x) ?

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

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integral of 4xarctan(x)

\int 4 x arctan (x) d x

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

Solution steps

\int 4 x arctan (x) d x

Take the constant out:

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

= 4 \cdot \int x arctan (x) d x

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

x \to 9 lim (\frac{1}{x - 9})

\frac{d}{d x} ((x + \frac{2}{x}) (7 - 2 x^{3}))

\frac{1}{y} \frac{d y}{d x} = e^{x} + ln (y)

(8 - r^{2}) d r = r^{3} sin (θ) d θ, r (0) = 2

\int 4 x^{3} - 6 x^{2} + 3 d x

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