Evaluate the integral: I = integral arctan x dx

Evaluate the integral: $$ I = \int \arctan x\,dx. $$

Explanation of Method

Treating $\arctan x$ as a product $\arctan x\cdot1$ and applying integration by parts with $u=\arctan x$, $dv=dx$ leaves a leftover integral resolved by recognizing the numerator as half the derivative of the denominator.

Choose $u=\arctan x$, $dv=dx$.
Differentiate $u$ and integrate $dv$.
Evaluate the leftover integral via substitution.

Step 1 · Set up integration by parts

Let:

$$ u = \arctan x \quad \Rightarrow \quad du = \frac{dx}{1+x^2}, \qquad dv = dx \quad \Rightarrow \quad v = x. $$

Step 2 · Apply the formula

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

Step 3 · Evaluate the leftover integral

Let $u_1=1+x^2$, so $du_1=2x\,dx$:

$$ \int \frac{x}{1+x^2}\,dx = \frac12\int \frac{du_1}{u_1} = \frac12\ln|u_1| = \frac12\ln(1+x^2). $$

Step 4 · Combine

$$ I = x\arctan x - \frac12\ln(1+x^2) + C. $$

Summary Table

Quantity	Value
$u$	$\arctan x$
$dv$	$dx$
Antiderivative	$x\arctan x-\tfrac12\ln(1+x^2)$

Additional Notes

The integrand is defined for all real $x$.
This is the companion to Problem 173: both problems use the same $u=$ inverse-function, $dv=dx$ strategy, but the leftover integral resolves to a square root here instead of a logarithm because $\arctan x$'s derivative $1/(1+x^2)$ doesn't have the same radical structure as $\arcsin x$'s.
The result can be checked by differentiating $x\arctan x-\tfrac12\ln(1+x^2)$ and confirming it returns $\arctan x$.

Result

The solution to the integral is:

$$ I = x\arctan x - \frac12\ln(1+x^2) + C. $$