Animated Solution for Mathematics - Indefinite Integration: Let $f(x)=\int \frac{dx}{x^{(2/3)}+2x^{(1/2)}}$ be such that $f(0)=-26+24{\log }_e(2)$. If $f(1)=a+b{\log }_e(3)$, where $a,b\in \mathbb{Z}$, then $a+b$ is equal to :

Welcome, fellow traveler on the path to JEE excellence. Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of fractional exponents.

We are looking at the integral:

When you see expressions like this, it is natural to feel a moment of hesitation. The fractional powers $x^{\frac{2}{3}}$ and $x^{\frac{1}{2}}$ are not just numbers; they are barriers to the standard power rule of integration. But in the world of calculus, every barrier is just an invitation to perform a clever transformation.

How do we clear these denominators? We need a substitution that turns these fractional powers into clean, beautiful integers.

We look at the denominators of our exponents: 3 and 2. To make both exponents integers, we need a power of $x$ that is a multiple of both 3 and 2. The least common multiple (LCM) of 3 and 2 is 6.

This leads us to the brilliant substitution: $x=u^6$. By choosing $u^6$, we ensure that:

Suddenly, the complexity vanishes, and we are left with a simple polynomial denominator: $u^4+2u^3$.

We cannot simply change $x$ to $u$ without accounting for the differential $dx$. If $x=u^6$, then the derivative of $x$ with respect to $u$ is:

Now, let us assemble our new integral. Substituting our terms, we get:

Look at the denominator again. We can factor out $u^3$ to get $u^3(u+2)$. This allows us to cancel $u^3$ from the numerator and denominator, simplifying our integral to:

This is the moment where the problem shifts from 'impossible' to 'manageable.'

We are now facing the integral of an improper rational function: $\frac{u^2}{u+2}$. Because the degree of the numerator (2) is greater than the degree of the denominator (1), we must simplify.

We know that $u^2-4$ factors into $(u-2)(u+2)$. So, let us rewrite $u^2$ as $(u^2-4)+4$. This gives us:

Now, the integration becomes a simple task of term-by-term calculus. We integrate $u$, $-2$, and $\frac{4}{u+2}$ separately:

Distributing the 6, we arrive at:

We are almost at the finish line. We must return to our original variable $x$. Since $x=u^6$, we have $u=x^{\frac{1}{6}}$. Substituting this back, our function becomes:

The problem provides us with a boundary condition: $f(0)=-26+24\ln (2)$. By plugging in $x=0$, the terms involving $x$ vanish, leaving us with:

The logarithmic terms cancel out perfectly, revealing that $C=-26$. Finally, we evaluate $f(1)$:

This simplifies to $3-12+24\ln (3)-26$, which is $-35+24\ln (3)$. Comparing this to $a+b\ln (3)$, we find $a=-35$ and $b=24$.

The final result is: