Evaluate the integral: I = integral (arctan³x)/(1+x²) dx

Evaluate the integral: $$ I = \int \frac{\arctan^3x}{1+x^2}\,dx. $$

Explanation of Method

The factor $1/(1+x^2)$ is exactly the derivative of $\arctan x$, so substituting $u=\arctan x$ reduces the integral to a plain power-rule form $\int u^3\,du$.

Substitute $u=\arctan x$.
Rewrite the integral in terms of $u$.
Apply the power rule for integration.
Substitute back and add the constant of integration.

Step 1 · Set up the substitution

Let:

$$ u = \arctan x. $$

Then:

$$ du = \frac{dx}{1+x^2}. $$

Step 2 · Rewrite the integral

Substituting into the integral:

$$ I = \int u^3\,du. $$

Step 3 · Apply the power rule

Integrating $u^3$:

$$ I = \frac{u^4}{4} + C. $$

Step 4 · Substitute back

Substituting back $u=\arctan x$:

$$ I = \frac{\arctan^4x}{4} + C. $$

Summary Table

Quantity	Value
Substitution	$u = \arctan x$
$dx/(1+x^2)$ in terms of $du$	$du$
Antiderivative	$u^4/4$

Additional Notes

The integrand is defined for all real $x$, since $1+x^2>0$ always.
This is the same $u=\arctan x$ substitution as Problem 107, but here the outer function is a power rather than a logarithm — any power $\arctan^nx/(1+x^2)$ reduces the same way.
The result can be checked by differentiating $\arctan^4x/4$ with the chain rule and confirming it returns $\dfrac{\arctan^3x}{1+x^2}$.

Result

The solution to the integral is:

$$ I = \frac{\arctan^4x}{4} + C. $$