Question

(-10)/(5-3x), dx find the integral

Answer 1

u-substitution works well here.

Answer 2

yeah but i get stuck on replacing it im up to where du=-3dx

Answer 3

OK, so it's -10/-3(u) du. The -3 can just go in the denominator with u.

Answer 4

so then i can put the 10/3 out since its a constant and ill be left with the integral of 1/u du which if im not mistaken its ln(f(x) +c

Answer 5

Yup.

Answer 6

so is the answer suppose to be 10/3 ln(5-3x)

Answer 7

Then derive to test.

Answer 8

im not taht good a t taking derivatives but i got something else 1/(5-3x)(-3)

Answer 9

10/3 ln(5-3x)+c

Answer 10

\[\int-\frac{10}{5-3x}~dx\\
\int\frac{10}{-(5-3x)}~dx\\
\int\frac{10}{3x-5}~dx\\
10\int\frac{1}{3x-5}~dx\]
Let \(u=3x-5\). Then, \(du=3~dx\), or \(\dfrac{1}{3}~du=dx\).
\[10\int\frac{1}{u}~\left(\frac{1}{3}~du\right)\\
\frac{10}{3}\int\frac{du}{u}\]

Answer 11

A much easier/faster method would be using natural log. It will take you about 2-3 lines of working.
\[\int\limits -\frac{10}{5-3x}dx=\int\limits \frac{10}{3x-5}dx\]
Get rid of that negative sign by putting it in the denominator.
\[=10\int\limits \frac{1}{3x-5}dx\]
Take out the 10 since it can deal by itself.
\[=10[\frac{1}{3}\ln(3x-5)]\]
When you anti-differentiate, you take a guess that it will be ln(3x-5)<= [the 3x-5 comes from the denominator] When you differentiate ln(3x-5) you have to also differentiate the inside of ln, which means a 3 comes out when you differentiate 3x. But you don't have a 3 in the integral, so you take the 3 out by putting 1/3 next to ln(3x-5) instead.
Now you can simply that by putting 10 back into the expression.

NOTE: If you have not learnt natural logarithms, then I suggest you go with what others have told you.

Answer 12

That's the exact same thing as u-substitution. The reason for the u-substitution is more for bookkeeping purposes than anything else in this case, but for a more complex integral it's good to have practice with simple u-substitutions.