 Sattelhofsk

2022-06-19

Where does this fraction come from in this integral?
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:
$-\frac{3}{4}v-\frac{150}{4}ln\left(200-4v\right)+C$
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. feaguelaBapzo

Expert

Using $u$-substitution, let $u=200-4v$:
$=\int \frac{0.75\left(200-u\right)}{u}\left(-\frac{1}{4}\mathrm{d}u\right)$
$=\int \left(\frac{3}{16}-\frac{150}{4u}\right)\mathrm{d}u$
$=\frac{3}{16}u-\frac{150}{4}\mathrm{ln}|u|+C$
$=\frac{3}{16}\left(200-4v\right)-\frac{150}{4}\mathrm{ln}|200-4v|+C$
$=\frac{600}{16}-\frac{3}{4}v-\frac{150}{4}\mathrm{ln}|200-4v|+C$
$=-\frac{3}{4}v-\frac{150}{4}\mathrm{ln}|200-4v|+C$
In the last step, the constant $\frac{600}{16}$ is absorbed into $C$
That is probably why you thought $u$-substitution did not work. Brenden Tran

Expert

$3\int \frac{x}{200-4x}dx=3\int \left[-\frac{25}{2\left(x-50\right)}-\frac{1}{4}\right]dx=-\frac{75}{2}\int \frac{1}{x-50}dx-\frac{3}{4}\int dx=-\frac{75}{2}\mathrm{ln}\left(x-50\right)-\frac{3}{4}x+c$
The first step is called "long division". This is the exact solution. (Note : I used $x$ for $v$ as expression.)

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