Q.Evaluate the definite integral $\int_{0}^{1}\frac{dx}{\sqrt{1+x}-\sqrt{x}}$

We start with

The denominator is a complex number of the form $2 + e^{i\theta}$. Our goal is to find a relation between $x$ and $y$ that doesn’t involve $\theta$.

Why rationalize?

When a complex number appears in the denominator, we can multiply numerator and denominator by its conjugate to separate real and imaginary parts. This is the same trick used for expressions like $\frac{1}{a+ib}$. Here, the conjugate of $2 + \cos\theta + i\sin\theta$ is $2 + \cos\theta - i\sin\theta$.

Step-by-step

1. Multiply numerator and denominator by the conjugate

2. Simplify the denominator

(We used $\cos^2\theta + \sin^2\theta = 1$.)

3. Separate real and imaginary parts

So

4. Eliminate $\theta$ We want $(x-1)(x-3)$. Let’s express $x$ in a more convenient form.

Compute $x-1$ and $x-3$:

Factor:

5. Multiply them $$$$

Concept and Intuition

A $0/0$ indeterminate form involving a square root is almost always resolved by rationalizing (multiplying by the conjugate, or equivalently recognizing $\sqrt x - 1 = \dfrac{x-1}{\sqrt x+1}$) so that the offending $(x-1)$ factor becomes explicit and cancels with the same factor in the denominator.

Step-by-Step Solution

1. Check the form at $x=1$: numerator $(9-1)(1-1)=0$; denominator $3+2-5=0$ — indeed $0/0$.
2. Factor the denominator: $3x^2+2x-5=(3x+5)(x-1)$ (verify: $3x^2-3x+5x-5=3x^2+2x-5$ ✓).
3. Rewrite $\sqrt x-1=\dfrac{(\sqrt x-1)(\sqrt x+1)}{\sqrt x+1}=\dfrac{x-1}{\sqrt x+1}$.
4. So the limit becomes …

Concept and Intuition

For a $\tfrac{0}{0}$-type limit built from nested square roots, the cleanest approach is the small-$t$ approximation $\sqrt{1+t}\approx 1+\dfrac{t}{2}$ (from the binomial series, valid to first order as $t\to0$), applied from the inside out. Since we only need the coefficient of $y^4$ in the numerator, first-order expansions suffice — no need for full Taylor series or L'Hôpital's rule (which would require four differentiations here).

Step-by-Step Solution

1. As $y\to0$, $\sqrt{1+y^4}\approx 1+\dfrac{y^4}{2}$ (first-order binomial expansion, $t=y^4\to0$).
2. So $1+\sqrt{1+y^4}\approx 2+\dfrac{y^4}{2}$.
3. Factor out $2$: $\sqrt{2+\tfrac{y^4}{2}}=\sqrt2\,\sqrt{1+\dfrac{y^4}{4}}$.
4. Expand again: $\sqrt{1+\dfrac{y^4}{4}}\approx 1+\dfrac{y^4}{8}$, so $\sqrt2\sqrt{1+\tfrac{y^4}{4}}\approx \sqrt2+\dfrac{\sqrt2}{8}y^4$.
5. Numerator: $\sqrt{1+\sqrt{1+y^4}}-\sqrt2 \approx \dfrac{\sqrt2}{8}y^4$. …

Concept and Intuition

When both numerator and denominator vanish at the limit point, replace $\sin x,\tan x$ by their Maclaurin series and keep terms up to the lowest surviving order — everything else washes out in the limit. For a difference of square roots like $\sqrt{f(x)}-\sqrt{g(x)}$, it helps to write $f,g$ as $3(1+u)$, $3(1+v)$ and expand $\sqrt{1+u}\approx 1+\tfrac{u}{2}$.

Step-by-Step Solution

1. Numerator: using $\sin x = x-\dfrac{x^3}{6}+\cdots$, $\tan x = x+\dfrac{x^3}{3}+\cdots$, $\tan^3 x = x^3+\cdots$:

As $x\to0$ this behaves like $6x$.

2. Denominator terms: $f(x)=x^2+2\sin x+\tan x+3 = 3 + (2x+x) + x^2 + O(x^3) = 3+3x+x^2+O(x^3)$.

$g(x)=\sin^2x-2\tan x-x+3 = 3 + (-2x-x) + x^2+O(x^3) = 3-3x+x^2+O(x^3)$. …

Concept and Intuition

At $x=2$: $\sqrt{1+8}=3$ and $\sqrt{3+6}=3$, so the numerator is $0$; also $x^3-8=0$. This is a genuine $0/0$ form, so L'Hôpital's rule (or equivalently rationalising) applies.

Step-by-Step Solution

1. Let $g(x)=\sqrt{1+4x}-\sqrt{3+3x}$, $h(x)=x^3-8$. Both $g(2)=0$, $h(2)=0$.
2. $g'(x)=\dfrac{4}{2\sqrt{1+4x}}-\dfrac{3}{2\sqrt{3+3x}}=\dfrac{2}{\sqrt{1+4x}}-\dfrac{1.5}{\sqrt{3+3x}}$.
3. At $x=2$: $g'(2)=\dfrac{2}{3}-\dfrac{1.5}{3}=\dfrac{0.5}{3}=\dfrac16$.
4. $h'(x)=3x^2$, so $h'(2)=12$. …