Animated Solution for Mathematics - Definite Integration: The value of ${\int }_{-1/\sqrt{2}}^{1/\sqrt{2}}{\left({\left(\frac{x+1}{x-1}\right)}^2+{\left(\frac{x-1}{x+1}\right)}^2-2\right)}^{1/2}dx$ is:

Welcome, fellow traveler on the path to JEE excellence. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of fractions and square roots.

You see an integral like:

Your instinct might be to panic. But I want you to take a deep breath. In mathematics, complexity is often just a mask for a simple, elegant truth waiting to be revealed.

Unmasking the Integrand

Let us look at the expression inside the square root. It is a classic algebraic structure.

Let us define $A=\frac{x+1}{x-1}$ and $B=\frac{x-1}{x+1}$. Notice the beautiful relationship between them:

This is not a coincidence; it is the key to the entire problem. The expression inside the root is $A^2+B^2-2$.

Since $A\cdot B=1$, we can rewrite this as $A^2+B^2-2AB$. Does that look familiar? It is the expansion of $(A-B{)}^2$!

So, our integrand simplifies to $\sqrt{(A-B{)}^2}$, which is simply $|A-B|$. We have already tamed the monster.

The Power of Symmetry

Now, let us simplify $A-B$. We have:

Finding a common denominator, we get:

Our integral is now $I={\int }_{-1/\sqrt{2}}^{1/\sqrt{2}}\left|\frac{4x}{x^2-1}\right|dx$.

Here is where we use the symmetry of the function. Because $f(x)=\left|\frac{4x}{x^2-1}\right|$ is an even function, the area under the curve from $-1/\sqrt{2}$ to $0$ is identical to the area from $0$ to $1/\sqrt{2}$.

We can rewrite our integral as:

This is the kind of strategic thinking that wins exams.

The Modulus Trap

We must be careful. In the interval $[0,1/\sqrt{2}]$, $x^2$ is less than $1/2$, which means $x^2-1$ is negative.

To remove the modulus, we must flip the sign: $|x^2-1|=1-x^2$. Our integral becomes:

To make this even cleaner, let us pull a $2$ out of the $4x$ to get:

Why did we do this? Because $2x$ is the derivative of $x^2$, and by extension, $-2x$ is the derivative of $1-x^2$. This is a perfect setup for substitution.

The Final Integration

Let $u=1-x^2$. Then $du=-2xdx$.

When $x=0$, $u=1$. When $x=1/\sqrt{2}$, $u=1-1/2=1/2$. Substituting these into our integral, we get:

The negative sign allows us to flip the limits:

The integral of $1/u$ is $\ln |u|$. Evaluating this, we get $4[\ln (1)-\ln (1/2)]$.

Since $\ln (1)=0$ and $-\ln (1/2)=\ln (2)$, we are left with $4\ln (2)$. Using the logarithmic power rule, $4\ln (2)=\ln (2^4)=\ln (16)$.

We started with a terrifying expression and ended with a clean, elegant $\ln (16)$. This is the joy of mathematics—taking the complex and finding the simple, logical path through it.