# Integral of x arctan x dx

## Homework Statement

$$\int x \arctan x \, dx$$

## The Attempt at a Solution

By parts,
$$u = \arctan x$$
$$dv = x dx$$
$$du = \frac{dx}{x^2+1}$$
$$v = \frac{x^2}{2}$$

$$\int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \frac{1}{2} \int \frac{x^2}{x^2+1} \, dx$$

Again...by parts

$$u = x^2$$
$$dv = \frac{dx}{x^2+1}$$
$$du = 2x dx$$
$$v = arc tan x$$

$$\int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \frac{x^2}{2}\arctan x - \int x \arctan x \, dx$$
I back to the beginning, what did wrogn?

$$\int x \arctan x \, dx = - \int x \arctan x \, dx$$

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$$\int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \frac{x^2}{2}\arctan x - \int x \arctan x \, dx$$

Add $$\int x \arctan x \, dx$$ to both sides, then solve for the integral, assuming your work is correct.

$$\int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \frac{x^2}{2}\arctan x - \int x \arctan x \, dx$$

Add $$\int x \arctan x \, dx$$ to both sides, then solve for the integral, assuming your work is correct.
you mean like this? is the same, i back to the beginign

$$\int x \arctan x \, dx +\int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \frac{x^2}{2}\arctan x - \int x \arctan x \, dx +\int x \arctan x \, dx$$

$$2\int x \arctan x \, dx = 0$$

$$- \frac{1}{2} \int \frac{x^2}{x^2+1} \, dx$$
Why use 'by parts' again? It would easier if you just add and subtract 1 from the numerator

why not try the substitution u=x^2+1 in that second integral...

for the integral x²/(x²+1)
you can rewrite it as (x² + 1 - 1)/(x²+1) => 1 - 1/(x²+1)

Gib Z
Homework Helper
umm hmm, that leaves a nice (x - arctan x) for you there.

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