Problem 281
Evaluate the integral: I = integral (sqrt x)/(1+ sqrt x)
Evaluate the integral: $$ I = \int \frac{\sqrt x}{1+\sqrt x}\,dx. $$
Explanation of Method
Substituting $u=\sqrt x$ clears the square root; the resulting improper rational function of $u$ splits by adding and subtracting $1$.
- Substitute $u=\sqrt x$.
- Simplify $\dfrac{u}{1+u}=1-\dfrac1{1+u}$.
- Integrate term by term.
- Back-substitute.
Step 1 · Substitute
Let $u=\sqrt x$, so $x=u^2$ and $dx=2u\,du$:
$$ I = \int\frac{u}{1+u}\cdot2u\,du = 2\int\frac{u^2}{1+u}\,du. $$
Step 2 · Divide
$$ \frac{u^2}{1+u} = u-1+\frac1{1+u}. $$
Step 3 · Integrate and back-substitute
$$ 2\int\left(u-1+\frac1{1+u}\right)du = u^2-2u+2\ln|1+u|+C. $$
$$ I = x - 2\sqrt x + 2\ln\left(1+\sqrt x\right) + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $u-1$ | $u^2/2-u$ |
| $1/(1+u)$ | $\ln|1+u|$ |
Additional Notes
- The domain requires $x\geq0$.
- The result was verified by differentiation: $\dfrac{d}{dx}\left[x-2\sqrt x+2\ln(1+\sqrt x)\right] = 1-\dfrac1{\sqrt x}+\dfrac1{\sqrt x(1+\sqrt x)} = 1-\dfrac1{1+\sqrt x} = \dfrac{\sqrt x}{1+\sqrt x}$, matching the integrand.
Result
The solution to the integral is:
$$ I = x - 2\sqrt x + 2\ln\left(1+\sqrt x\right) + C. $$