Animated Solution for Mathematics - Indefinite Integration: Evaluate $\int \frac{x^{2}dx}{(a+bx{)}^2}$

The integral we are evaluating is:

When you first see this, your instinct might be to panic or try to expand the denominator, but let's pause and look at the structure. The denominator $(a+bx{)}^2$ is the bottleneck.

In calculus, whenever you see a linear expression trapped inside a power, it is a signal to use substitution. By defining $t=a+bx$, we are essentially changing our perspective, shifting the complexity from the denominator to the numerator.

Let's put our strategy into action. We set $t=a+bx$.

Now, we must be rigorous. We need to change the differential $dx$ to $dt$. Differentiating both sides, we get $bdx=dt$, which means $dx=\frac{dt}{b}$.

But we aren't done yet. We have an $x^2$ in the numerator. We must express $x$ in terms of $t$. From our substitution $t=a+bx$, we isolate $x$: $bx=t-a$, so $x=\frac{t-a}{b}$.

Substituting these into our integral, we get:

Let's clean this up. We have a constant $\frac{1}{b^2}$ from the squared term and another $\frac{1}{b}$ from the differential, giving us $\frac{1}{b^3}$ outside the integral.

Now, we expand the numerator: $(t-a{)}^2=t^2-2at+a^2$. Our integral becomes:

This is where the magic happens. Because the denominator is a single term, $t^2$, we can divide each term in the numerator by $t^2$ individually. This gives us:

Now, we integrate term by term. The integral of $1$ is $t$. The integral of $-\frac{2a}{t}$ is $-2a\ln |t|$. The integral of $\frac{a^2}{t^2}$ is $-\frac{a^2}{t}$.

Combining these, we get:

Finally, we revert to $x$ by substituting $t=a+bx$ back into the expression. The result is:

This journey shows that even the most daunting integrals can be tamed with the right substitution. Keep practicing, and you will find that calculus is not just about solving problems—it's about finding the hidden elegance in the math.