Animated Solution for Mathematics - Definite Integration: If ${\int }_0^1\frac{1}{\sqrt{3+x}+\sqrt{1+x}}dx=a+b\sqrt{2}+c\sqrt{3}$, where $a,b,c$ are rational numbers, then $2a+3b-4c$ is equal to :

A Journey Through the Integral

Welcome, fellow traveler on the road to JEE excellence! Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of square roots.

You might be tempted to reach for a complex substitution, but let's pause and look at the structure of the integrand:

Whenever you see a sum of square roots in a denominator, your mathematical intuition should immediately signal the Conjugate Strategy. The conjugate is the key that unlocks the door to a much simpler world.

By multiplying the numerator and the denominator by $\sqrt{3+x}-\sqrt{1+x}$, we transform the denominator using the difference of squares identity: $(A+B)(A-B)=A^2-B^2$.

Observe what happens: the denominator becomes $(\sqrt{3+x}{)}^2-(\sqrt{1+x}{)}^2$, which simplifies beautifully to $(3+x)-(1+x)=2$. The variable $x$ vanishes from the denominator entirely!

We are left with a much friendlier integral:

Now that we have separated the terms, we can apply the power rule for integration. Remember, $\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}$.

Applying this to our terms, we get:

Simplifying the coefficients, we arrive at:

At $x=1$, we have $\frac{1}{3}[4^{3/2}-2^{3/2}]$. Since $4^{3/2}=(2^2{)}^{3/2}=2^3=8$ and $2^{3/2}=2\sqrt{2}$, this becomes $\frac{1}{3}[8-2\sqrt{2}]$.

At $x=0$, we have $\frac{1}{3}[3^{3/2}-1^{3/2}]$. Since $3^{3/2}=3\sqrt{3}$ and $1^{3/2}=1$, this becomes $\frac{1}{3}[3\sqrt{3}-1]$.

Subtracting these, we get:

By comparing this to $a+b\sqrt{2}+c\sqrt{3}$, we identify $a=3$, $b=-\frac{2}{3}$, and $c=-1$.

Finally, calculating $2a+3b-4c$ gives us:

See how the complexity melted away? You didn't need a sledgehammer; you just needed the right tool. Keep practicing, keep questioning, and keep falling in love with the elegance of the process!