Animated Solution for Mathematics - Definite Integration: Evaluate ${\int }_2^3\frac{2x^5+x^4-2x^3+2x^2+1}{\left(x^2+1\right)\left(x^4-1\right)}dx$

Welcome, warriors of JEE Advanced. Today, we stand before a problem that, at first glance, looks like a chaotic mess of polynomials. We are asked to evaluate the definite integral:

When you see a high-degree rational function like this, do not panic. In the world of competitive mathematics, intimidation is the first trap. Our job is to peel back the layers of this expression to reveal the elegant structure hidden underneath.

Decoding the Denominator

Before we even think about integration, we must simplify the landscape. Look at the denominator: $\left(x^2+1\right)\left(x^4-1\right)$.

That $x^4-1$ is a classic difference of squares, $a^2-b^2=(a-b)(a+b)$. Thus, $x^4-1=(x^2-1)(x^2+1)$.

When we multiply this by the existing $(x^2+1)$, our denominator becomes $(x^2+1{)}^2(x^2-1)$. We have transformed a complex product into a structured form that we can actually work with.

The Surgical Strike on the Numerator

Now, we turn our attention to the numerator: $2x^5+x^4-2x^3+2x^2+1$. If we try to divide this directly, we will drown in algebra.

Instead, let us use a bit of intuition to create terms that look like our denominator factors. Let us group the terms strategically: $(2x^5-2x^3)+(x^4+2x^2+1)$.

Look at the first group: $2x^3(x^2-1)$. Look at the second group: $(x^2+1{)}^2$. We have successfully rewritten the numerator as $2x^3(x^2-1)+(x^2+1{)}^2$.

The Great Split

Now, we bring the numerator and denominator together. Our integral becomes:

We can split this into two separate integrals:

Notice the magic? The terms cancel out perfectly! We are left with two manageable integrals:

Solving the Components

Let us solve the second part, $I_2={\int }_2^3\frac{1}{x^2-1}dx$, first. This is a standard form: $\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\ln \left|\frac{x-a}{x+a}\right|$.

With $a=1$, this becomes:

Evaluating this, we get:

Now for the first part, $I_1={\int }_2^3\frac{2x^3}{(x^2+1{)}^2}dx$. We use substitution. Let $t=x^2+1$, so $dt=2xdx$.

We rewrite $2x^3$ as $x^2(2x)$, which is $(t-1)dt$. The limits change from $x=2,3$ to $t=5,10$. The integral becomes:

Integrating this gives:

The Final Synthesis

Combining our results, $I=I_1+I_2=(\ln 2-\frac{1}{10})+\frac{1}{2}\ln \left(\frac{3}{2}\right)$.

Expanding the log term, we get $\ln 2-\frac{1}{10}+\frac{1}{2}\ln 3-\frac{1}{2}\ln 2$. Simplifying the $\ln 2$ terms, we arrive at:

We have conquered the beast!