Where does this fraction come from in this integral?

Question

So you have the integral: $$\int\frac{3v}{200 - 4v} dv$$ I tried to do $u$-substitution at first with $u = 200 - 4v$, but I could not get the correct answer which is: $$-\frac{3}{4}v - \frac{150}{4}ln(200-4v) + C$$ In the worked solution, they did not use a $u$-substitution. The first integral becomes: $$\int -\frac{3}{4} + \frac{150}{200 - 4v} dv$$ And I cannot see what technique they used to get that. I worked out that if you actually add the 2 fractions you end up back at the first integral, but I do not see how they worked out that is the way it should be re-arranged. I also don't understand why my $u$-substitution didn't work. Should a $u$-substitution have worked? I'm still trying to get my head around this integrating of fractions.

Answer 1

$3\int \frac{x}{200-4x}dx = 3\int[ -\frac{25}{2(x-50)} - \frac{1}{4} ]dx = -\frac{75}{2}\int \frac{1}{x-50}dx - \frac{3}{4}\int dx = -\frac{75}{2}\ln(x-50) -\frac{3}{4}x + c$

The first step is called "long division". This is the exact solution. (Note : I used $x$ for $v$ as expression.)

Answer 2

Using $u$-substitution, let $u=200-4v$, $\mathrm du=-4\ \mathrm dv$:

$$=\int\frac{0.75(200-u)}{u}\left(-\frac14\mathrm du\right)$$

$$=\int\left(\frac3{16}-\frac{150}{4u}\right)\mathrm du$$

$$=\frac3{16}u-\frac{150}4\ln|u|+C$$

$$=\frac3{16}(200-4v)-\frac{150}4\ln|200-4v|+C$$

$$=\frac{600}{16}-\frac34v-\frac{150}4\ln|200-4v|+C$$

$$=-\frac34v-\frac{150}4\ln|200-4v|+C$$

In the last step, the constant $\frac{600}{16}$ is absorbed into $C$.

That is probably why you thought $u$-substitution did not work.

Answer 3

we have $$\frac{3v}{200-4v}=\frac{3}{4}\cdot \frac{4v}{200-4v}=-\frac{3}{4}\cdot \frac{-4v}{200-4v}=-\frac{3}{4}\cdot \left(\frac{200-4v-200}{200-4v}\right)=-\frac{3}{4}\left(1-\frac{200}{200-4v}\right)$$