Sattelhofsk

Answered

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:

$-\frac{3}{4}v-\frac{150}{4}ln(200-4v)+C$

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.

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 & Explanation

feaguelaBapzo

Expert

2022-06-20Added 9 answers

Using $u$-substitution, let $u=200-4v$:

$=\int \frac{0.75(200-u)}{u}(-\frac{1}{4}\mathrm{d}u)$

$=\int (\frac{3}{16}-\frac{150}{4u})\mathrm{d}u$

$=\frac{3}{16}u-\frac{150}{4}\mathrm{ln}|u|+C$

$=\frac{3}{16}(200-4v)-\frac{150}{4}\mathrm{ln}|200-4v|+C$

$=\frac{600}{16}-\frac{3}{4}v-\frac{150}{4}\mathrm{ln}|200-4v|+C$

$=-\frac{3}{4}v-\frac{150}{4}\mathrm{ln}|200-4v|+C$

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

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

$=\int \frac{0.75(200-u)}{u}(-\frac{1}{4}\mathrm{d}u)$

$=\int (\frac{3}{16}-\frac{150}{4u})\mathrm{d}u$

$=\frac{3}{16}u-\frac{150}{4}\mathrm{ln}|u|+C$

$=\frac{3}{16}(200-4v)-\frac{150}{4}\mathrm{ln}|200-4v|+C$

$=\frac{600}{16}-\frac{3}{4}v-\frac{150}{4}\mathrm{ln}|200-4v|+C$

$=-\frac{3}{4}v-\frac{150}{4}\mathrm{ln}|200-4v|+C$

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

2022-06-21Added 9 answers

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

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

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

Most Popular Questions