Evaluate the integral: I = integral (sqrt x)/((x+1)²) dx

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

Step 1 · Substitute

Let $u=\sqrt x$, so $x=u^2$ and $dx=2u\,du$:

$$ I = \int\frac u{(u^2+1)^2}\cdot2u\,du = 2\int\frac{u^2}{(u^2+1)^2}\,du. $$

Step 2 · Rewrite and integrate

$$ \frac{u^2}{(u^2+1)^2} = \frac1{u^2+1} - \frac1{(u^2+1)^2}. $$

Using $\displaystyle\int\frac{du}{(u^2+1)^2}=\frac u{2(u^2+1)}+\frac12\arctan u$:

$$ 2\int\frac{u^2}{(u^2+1)^2}\,du = 2\arctan u - \left[\frac u{u^2+1}+\arctan u\right] = \arctan u - \frac u{u^2+1}. $$

Step 3 · Back-substitute

$$ I = \arctan\sqrt x - \frac{\sqrt x}{x+1} + C. $$

Summary Table

Additional Notes

• The domain requires $x\geq0$.
• The result was verified by differentiation: combining the derivatives of $\arctan\sqrt x$ and $-\sqrt x/(x+1)$ over the common denominator $2\sqrt x(x+1)^2$ gives numerator $2x$, so the derivative is $\dfrac{2x}{2\sqrt x(x+1)^2}=\dfrac{\sqrt x}{(x+1)^2}$, matching the integrand.