Evaluate the integral: I = integral dx/(1+ sqrt x+
Evaluate the integral: $$ I = \int \frac{dx}{1+\sqrt x+\sqrt{1+x}}. $$
Explanation of Method
After substituting $u=\sqrt x$ and $v=\sqrt{1+u^2}$, the key move is rationalizing $\dfrac{2u}{1+u+v}$ by multiplying by the conjugate $1+u-v$. Since $(1+u)^2-v^2=2u$ exactly, this collapses the entire fraction to the simple expression $1+u-v$, with no remaining denominator at all.
- Substitute $u=\sqrt x$.
- Rationalize by the conjugate $1+u-\sqrt{1+u^2}$.
- Integrate the simplified expression.
- Back-substitute.
Step 1 · Substitute
Let $u=\sqrt x$, so $x=u^2$, $dx=2u\,du$, and $\sqrt{1+x}=\sqrt{1+u^2}$:
$$ I = \int\frac{2u\,du}{1+u+\sqrt{1+u^2}}. $$
Step 2 · Rationalize
Multiplying numerator and denominator by $1+u-\sqrt{1+u^2}$, and using $(1+u)^2-\left(1+u^2\right)=2u$:
$$ \frac{2u}{1+u+\sqrt{1+u^2}} = \frac{2u\left(1+u-\sqrt{1+u^2}\right)}{(1+u)^2-\left(1+u^2\right)} = \frac{2u\left(1+u-\sqrt{1+u^2}\right)}{2u} = 1+u-\sqrt{1+u^2}. $$
The denominator vanishes entirely — the integrand becomes a simple algebraic expression with no fraction at all.
Step 3 · Integrate
Using $\displaystyle\int\sqrt{1+u^2}\,du = \frac u2\sqrt{1+u^2}+\frac12\ln\left(u+\sqrt{1+u^2}\right)$:
$$ I = \int\left(1+u-\sqrt{1+u^2}\right)du = u + \frac{u^2}2 - \frac u2\sqrt{1+u^2} - \frac12\ln\left(u+\sqrt{1+u^2}\right) + C. $$
Step 4 · Back-substitute
$$ I = \sqrt x + \frac x2 - \frac{\sqrt x}2\sqrt{1+x} - \frac12\ln\left(\sqrt x+\sqrt{1+x}\right) + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $1+u$ | $u+u^2/2$ |
| $-\sqrt{1+u^2}$ | $-\tfrac u2\sqrt{1+u^2}-\tfrac12\ln\left(u+\sqrt{1+u^2}\right)$ |
Additional Notes
- Major rewrite: the source's attempt got tangled in a partial rationalization, abandoned the calculation midway ("this becomes complex... requires advanced techniques"), and asserted a final boxed answer without real justification. Differentiating that proposed answer at $x=1$ gives $\approx-0.914$, while the true integrand value is $1/(2+\sqrt2)\approx0.293$ — not even the right sign, confirming it was wrong. The key simplification the source missed is that multiplying by the conjugate $1+u-\sqrt{1+u^2}$ makes the denominator collapse to exactly $2u$, canceling the $u$ in the numerator completely and leaving no fraction to integrate at all.
- The domain requires $x\geq0$.
- The result was verified two ways: symbolically, by confirming $(1+u)^2-(1+u^2)=2u$ identically; and numerically, at $x=1$ ($u=1$), where $\tfrac{dI}{du}=1+1-\sqrt2=2-\sqrt2\approx0.586$ and $\tfrac{du}{dx}=\tfrac12$, giving $\tfrac{dI}{dx}\approx0.293$, matching the integrand $1/(2+\sqrt2)\approx0.293$ exactly.
Result
The solution to the integral is: