Evaluate the integral: I = integral arctan sqrt x dx

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

Step 1 · Set up integration by parts

Let:

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

Applying the formula:

$$ I = x\arctan\sqrt x - \frac12\int \frac{\sqrt x}{1+x}\,dx. $$

Step 2 · Evaluate the leftover integral

Let $w=\sqrt x$, so $x=w^2$, $dx=2w\,dw$:

$$ \int \frac{\sqrt x}{1+x}\,dx = \int \frac{w}{1+w^2}\cdot2w\,dw = 2\int \frac{w^2}{1+w^2}\,dw. $$

Step 3 · Split and integrate

Writing $\dfrac{w^2}{1+w^2}=1-\dfrac{1}{1+w^2}$:

$$ 2\int \left(1-\frac{1}{1+w^2}\right) dw = 2\left(w - \arctan w\right) = 2\sqrt x - 2\arctan\sqrt x. $$

Step 4 · Combine

Substituting back into the expression for $I$:

$$ I = x\arctan\sqrt x - \frac12\left(2\sqrt x - 2\arctan\sqrt x\right) + C = x\arctan\sqrt x - \sqrt x + \arctan\sqrt x + C. $$

Combining the two $\arctan\sqrt x$ terms:

$$ I = (x+1)\arctan\sqrt x - \sqrt x + C. $$

Summary Table

Additional Notes

• The domain requires $x\geq0$, where $\sqrt x$ is real.
• Substituting $w=\sqrt x$ directly into the leftover integral (rather than a chain of intermediate substitutions) reveals it as the same $\int w^2/(1+w^2)\,dw$ pattern used in Problem 109 — recognizing this shortcut avoids unnecessary algebra.
• The result can be checked by differentiating $(x+1)\arctan\sqrt x-\sqrt x$: this gives $\arctan\sqrt x+\tfrac{x+1}{2\sqrt x(1+x)}-\tfrac{1}{2\sqrt x}=\arctan\sqrt x$, matching the integrand.