Animated Solution for Mathematics - Indefinite Integration: Let $f(x)=\int \frac{2x}{(x^2+1)(x^2+3)}dx$. If $f(3)=\frac{1}{2}({\log }_{e}5-{\log }_{e}6)$, then $f(4)$ is equal to

Welcome, future engineer! Today, we are going to unravel a beautiful integral that might look intimidating at first glance, but hides a core of elegant simplicity. The problem asks us to evaluate $f(4)$ given the function:

When you look at this integral, do not panic. Instead, look for the hidden symmetry. Notice the $2x$ in the numerator and the $x^2$ terms in the denominator; this is a classic 'lock and key' mechanism where the derivative of $x^2$ is $2x$.

This is a massive hint that we should use the substitution $t=x^2$. By setting $t=x^2$, we differentiate both sides to get $dt=2xdx$. Suddenly, the entire numerator $2xdx$ is replaced by $dt$, transforming the integral into:

Now that we have $\int \frac{dt}{(t+1)(t+3)}$, we face a new challenge. We cannot integrate this product directly, so we employ the method of partial fractions. We want to express the integrand as a difference of two simpler fractions.

Notice that the difference between the factors $(t+3)$ and $(t+1)$ is exactly $2$. Therefore, we can write:

This step is crucial. If you take the common denominator of the right side, you get $\frac{(t+3)-(t+1)}{2(t+1)(t+3)}=\frac{2}{2(t+1)(t+3)}$, which simplifies back to the original expression. Just remember to keep that $\frac{1}{2}$ outside the integral to maintain the balance.

With the integral split, we can now integrate term by term. The integral of $\frac{1}{t+1}$ is $\ln |t+1|$, and the integral of $\frac{1}{t+3}$ is $\ln |t+3|$. Our function becomes:

Using the logarithmic property $\ln A-\ln B=\ln (\frac{A}{B})$, we simplify this to $f(x)=\frac{1}{2}\ln \left(\frac{t+1}{t+3}\right)+C$. Back-substituting $t=x^2$, we return to our original variable:

To find $C$, we use the boundary condition $f(3)=\frac{1}{2}(\ln 5-\ln 6)$. Plugging $x=3$ into our formula:

Comparing this to the given $f(3)$, it is clear that $C=0$.

We have our complete function: $f(x)=\frac{1}{2}\ln \left(\frac{x^2+1}{x^2+3}\right)$. Now, finding $f(4)$ is just a matter of arithmetic. Substitute $x=4$:

The final result is:

And there you have it! A complex integral conquered through substitution, partial fractions, and a bit of logarithmic elegance. Keep practicing, and these steps will become second nature to you.