Animated Solution for Mathematics - Definite Integration: The value of the integral ${\int }_1^2\left(\frac{t^4+1}{t^6+1}\right)dt$ is

When you first look at the integral

it is perfectly normal to feel a moment of hesitation. The powers are high, the fraction looks rigid, and there is no obvious substitution staring you in the face.

But in the world of JEE Advanced, we do not fear complexity; we dismantle it. Let us peel back the layers of this problem together.

The first step in any rational integration is to understand the structure of the denominator. We have $t^6+1$.

If you stare at this long enough, you might see the ghost of an algebraic identity. Recall the sum of cubes: $a^3+b^3=(a+b)(a^2-ab+b^2)$.

If we let $a=t^2$ and $b=1$, then $t^6+1$ becomes $(t^2{)}^3+1^3$. This factors beautifully into:

Suddenly, the denominator is no longer a monolithic block; it is a product of two distinct, manageable pieces.

Now, we turn our attention to the numerator, $t^4+1$. We need this to 'talk' to the denominator.

Look at the factor $(t^4-t^2+1)$ we just uncovered. It is almost identical to our numerator!

To make them match, we perform a classic JEE maneuver: add and subtract $t^2$. We rewrite the numerator as $(t^4-t^2+1)+t^2$. This is not just algebra; it is a strategic decision to force a cancellation.

By substituting this new numerator back into our fraction, we get:

Now, we split the fraction at the plus sign. The first term becomes:

The second term becomes:

We have transformed one intimidating integral into two friendly ones.

The first integral, $\int \frac{1}{t^2+1}dt$, is a standard result: ${\tan }^{-1}t$.

The second integral, $\int \frac{t^2}{t^6+1}dt$, requires a gentle nudge. Notice that the derivative of $t^3$ is $3t^2$.

This is our signal! We set $u=t^3$, which means $du=3t^{2}dt$, or $t^{2}dt=\frac{1}{3}du$. The integral becomes:

Substituting back, we get $\frac{1}{3}{\tan }^{-1}(t^3)$.

We combine our results:

Evaluating at the upper limit $t=2$, we get ${\tan }^{-1}2+\frac{1}{3}{\tan }^{-1}8$.

Evaluating at the lower limit $t=1$, we get:

Subtracting the lower from the upper, we arrive at our final, elegant answer:

You have not just solved a problem; you have mastered the art of algebraic decomposition. Keep this mindset, and no integral will ever be too daunting again.