Evaluate the integral: I = integral (arctan
Evaluate the integral: $$ I = \int \frac{\arctan\sqrt{x}}{\sqrt{x}}\cdot\frac{1}{1+x}\,dx. $$
Explanation of Method
Substituting $u=\sqrt x$ removes the square roots and turns the integral into one involving $\arctan u/(1+u^2)$, which is itself a standard derivative-matching pattern, resolved by a second substitution $v=\arctan u$.
- Substitute $u=\sqrt x$ to clear the radicals.
- Simplify to a $\arctan u/(1+u^2)$ form.
- Substitute $v=\arctan u$ and integrate.
- Substitute back and add the constant of integration.
Step 1 · Substitute $u=\sqrt x$
Let $u=\sqrt x$, so $x=u^2$ and:
$$ dx = 2u\,du. $$
Also, $\arctan\sqrt x = \arctan u$, $\dfrac{1}{\sqrt x}=\dfrac1u$, and $\dfrac{1}{1+x}=\dfrac{1}{1+u^2}$. Substituting:
$$ I = \int \frac{\arctan u}{u}\cdot\frac{1}{1+u^2}\cdot 2u\,du. $$
Step 2 · Simplify
The factors of $u$ cancel:
$$ I = 2\int \frac{\arctan u}{1+u^2}\,du. $$
Step 3 · Substitute $v=\arctan u$
Let $v=\arctan u$, so $dv=\dfrac{du}{1+u^2}$. Then:
$$ I = 2\int v\,dv = 2\cdot\frac{v^2}{2} + C = v^2 + C. $$
Step 4 · Substitute back
Since $v=\arctan u$ and $u=\sqrt x$, substituting back:
$$ I = \left(\arctan\sqrt x\right)^2 + C. $$
Summary Table
| Substitution | Result |
|---|---|
| $u=\sqrt x$ | clears the radicals |
| $v=\arctan u$ | reduces to $2\int v\,dv$ |
| Antiderivative | $v^2 = \arctan^2\sqrt x$ |
Additional Notes
- The domain requires $x\geq0$ for $\sqrt x$ to be real.
- This problem stacks two derivative-matching substitutions in sequence: first $u=\sqrt x$ clears the radical, then $v=\arctan u$ resolves the resulting arctangent pattern — a useful template whenever a radical and an inverse trig function appear together.
- The result can be checked by differentiating $\arctan^2\sqrt x$ with the chain rule: $\dfrac{d}{dx}(\arctan\sqrt x)^2 = 2\arctan\sqrt x\cdot\dfrac{1}{1+x}\cdot\dfrac{1}{2\sqrt x} = \dfrac{\arctan\sqrt x}{\sqrt x(1+x)}$, matching the original integrand.
Result
The solution to the integral is: