Animated Solution for Mathematics - Definite Integration: The integral $16{\int }_1^2\frac{dx}{x^3(x^2+2{)}^2}$ is equal to

Conquering the Integral

I know, looking at this integral, it feels like a wall of symbols designed to intimidate you: $16{\int }_1^2\frac{dx}{x^3(x^2+2{)}^2}$. It is heavy, it is dense, and it seems to defy standard integration techniques.

But in the world of JEE Advanced, every complex problem is just a puzzle waiting for the right key. Today, we are going to perform some algebraic surgery to unlock it.

The Hidden Structure

The denominator is our primary obstacle. We have $x^3(x^2+2{)}^2$. If we try to expand the square, we get a polynomial that is even harder to handle.

Instead, let us look for a hidden structure. Let us factor out $x^2$ from $(x^2+2{)}^2$. Remember, when $x^2$ comes out of a square, it becomes $x^4$.

So, $(x^2+2{)}^2=[x^2(1+\frac{2}{x^2}){]}^2=x^4(1+\frac{2}{x^2}{)}^2$. Now, substitute this back into our integral:

Suddenly, the landscape changes. We have a clear path forward.

The Substitution Strategy

Now, we need a substitution that simplifies the integrand. Look at the term $(1+\frac{2}{x^2})$.

If we let $t=1+\frac{2}{x^2}$, or $t=1+2x^{-2}$, let us see what happens when we differentiate. Using the power rule, $dt=-4x^{-3}dx$, which is $-\frac{4}{x^3}dx$.

This is brilliant! We have an $x^3$ in the denominator, and now we have a way to replace $\frac{dx}{x^3}$ with $-\frac{dt}{4}$.

The Transformation of Limits

Never forget the limits! When $x=1$, $t=1+\frac{2}{1^2}=3$. When $x=2$, $t=1+\frac{2}{2^2}=1+\frac{1}{2}=\frac{3}{2}$.

Our integral now spans from $t=3$ to $t=\frac{3}{2}$. We still have an $x^4$ in the denominator. We need to express it in terms of $t$.

Since $t=1+\frac{2}{x^2}$, we have $\frac{2}{x^2}=t-1$, so $\frac{1}{x^2}=\frac{t-1}{2}$. Squaring both sides, $\frac{1}{x^4}=(\frac{t-1}{2}{)}^2$.

The Final Integration

Putting it all together, the integral becomes:

The constants $16$, $\frac{1}{4}$ (from the square), and $\frac{1}{4}$ (from the $dt$ term) cancel out perfectly to $1$. We are left with:

This is a simple rational function! We can split it into three terms: ${\int }_{\frac{3}{2}}^3(1-\frac{2}{t}+\frac{1}{t^2})dt$. Integrating term by term, we get:

Evaluating at the upper limit $3$: $(3-2\ln 3-\frac{1}{3})=\frac{8}{3}-2\ln 3$.

Evaluating at the lower limit $\frac{3}{2}$: $(\frac{3}{2}-2\ln (\frac{3}{2})-\frac{2}{3})=\frac{5}{6}-2(\ln 3-\ln 2)=\frac{5}{6}-2\ln 3+2\ln 2$.

Subtracting the two, the $-2\ln 3$ terms cancel out, leaving us with $\frac{8}{3}-\frac{5}{6}-2\ln 2=\frac{11}{6}-\ln 4$.

And there it is! The elegance of the final result is the reward for your patience. Keep this trick in your toolkit: when you see a polynomial raised to a power in the denominator, look for that highest power to factor out. It is the key to unlocking the most intimidating integrals.