Problem 356
Evaluate the integral: I = integral (sqrt(1+ sqrt
Evaluate the integral: $$ I = \int \frac{\sqrt{1+\sqrt x}}{\sqrt x}\,dx. $$
Explanation of Method
The substitution $u=\sqrt x$ turns $dx/\sqrt x$ into $2\,du$, reducing the integral directly to a power-rule integral in $1+u$.
- Substitute $u=\sqrt x$.
- Substitute $v=1+u$.
- Integrate and back-substitute.
Step 1 · Substitute
Let $u=\sqrt x$, so $x=u^2$, $dx=2u\,du$, and $dx/\sqrt x=2\,du$:
$$ I = \int\sqrt{1+u}\cdot2\,du = 2\int\sqrt{1+u}\,du. $$
Step 2 · Integrate
$$ I = 2\cdot\frac23(1+u)^{3/2} + C = \frac43(1+u)^{3/2} + C. $$
Step 3 · Back-substitute
With $u=\sqrt x$:
Summary Table
| Substitution | Effect |
|---|---|
| $u=\sqrt x$ | $dx/\sqrt x\to2\,du$ |
Additional Notes
- The domain is $x>0$.
- The result was verified by differentiation at $x=4$: $\tfrac{d}{dx}\left[\tfrac43(1+\sqrt x)^{3/2}\right]=2(1+\sqrt x)^{1/2}\cdot\tfrac1{2\sqrt x}=\sqrt{1+\sqrt x}/\sqrt x$, matching the integrand exactly.
Result
The solution to the integral is:
$$ I = \frac43\left(1+\sqrt x\right)^{3/2} + C. $$