What is the integral of (x^2+3)arctan(x) ?

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

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integral of (x^2+3)arctan(x)

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

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

Solution steps

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

Expand

(x^{2} + 3) arctan (x) : x^{2} arctan (x) + 3 arctan (x)

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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