Animated Solution for Mathematics - Indefinite Integration: Let $f(x)=\int \frac{\sqrt{x}}{(1+x{)}^2}dx(x>0)$. Then $f(3)-f(1)$ is equal to :

Welcome, JEE warriors. Today, we are not just solving an integral; we are peeling back the layers of a mathematical onion.

When you first look at the function $f(x)=\int \frac{\sqrt{x}}{(1+x{)}^2}dx$, it is natural to feel a moment of hesitation. The denominator $(1+x{)}^2$ looks heavy, and the $\sqrt{x}$ in the numerator feels like it is begging for a change of variables.

In the world of JEE Advanced, this is where you stop, take a breath, and look for the hidden symmetry.

The Geometric Intuition

Why do we choose $\sqrt{x}=\tan \theta$? It is not a random guess. We are looking for a way to simplify the term $(1+x)$.

We know the fundamental identity $1+{\tan }^2\theta ={\sec }^2\theta$. If we set $\sqrt{x}=\tan \theta$, then $x={\tan }^2\theta$.

Suddenly, the denominator $(1+x{)}^2$ becomes $(1+{\tan }^2\theta {)}^2$, which is $({\sec }^2\theta {)}^2$, or ${\sec }^4\theta$. This is the 'Aha!' moment where we transform an algebraic expression into a trigonometric one ready to collapse under the weight of its own simplicity.

The Transformation

Now, we must be precise. If $x={\tan }^2\theta$, we cannot simply integrate with respect to $x$ anymore. We must find the differential $dx$.

Using the chain rule, the derivative of ${\tan }^2\theta$ with respect to $\theta$ is $2\tan \theta \cdot {\sec }^2\theta$. Therefore, $dx=2\tan \theta {\sec }^2\theta d\theta$.

When we substitute these into our integral, we get:

Look at that expression. It looks intimidating, but watch what happens when we clean it up. The denominator is ${\sec }^4\theta$, and the numerator has a ${\sec }^2\theta$ term. We can cancel them out to obtain:

The Elegance of Cancellation

This is where the beauty of trigonometry shines. We know that ${\tan }^2\theta =\frac{{\sin }^2\theta }{{\cos }^2\theta }$ and ${\sec }^2\theta =\frac{1}{{\cos }^2\theta }$.

When we divide ${\tan }^2\theta$ by ${\sec }^2\theta$, the ${\cos }^2\theta$ terms cancel out perfectly, leaving us with:

We have reduced a complex rational function to the integral of $2{\sin }^2\theta$. We know that $2{\sin }^2\theta =1-\cos 2\theta$.

Integrating this is trivial: the integral of $1$ is $\theta$, and the integral of $-\cos 2\theta$ is $-\frac{\sin 2\theta }{2}$. Thus, our indefinite integral is $\theta -\frac{\sin 2\theta }{2}+C$.

The Final Stretch

We are almost there. We must return to the world of $x$. Since $\sqrt{x}=\tan \theta$, we know $\theta ={\tan }^{-1}(\sqrt{x})$.

For the $\sin 2\theta$ term, we use the identity $\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$. Substituting $\sqrt{x}$ for $\tan \theta$, we get $\frac{2\sqrt{x}}{1+x}$.

Putting it all together, our function is:

Now, we evaluate $f(3)-f(1)$. For $x=3$:

For $x=1$:

Subtracting these, the constants $C$ vanish. We are left with $(\frac{\pi }{3}-\frac{\pi }{4})+\frac{1}{2}-\frac{\sqrt{3}}{4}$, which simplifies beautifully to the final answer:

See? The complexity was just a mask. By staying calm and trusting the identities, you have conquered the problem. Keep practicing, and soon, you will see these patterns instantly.