Thread: Integration of fractions to give ln quick help.

Hi everyone - very quick one here

Ill use f in bold do denote integration symbol because I havn't fully mastered the thing to enter formulae yet

Anyway I'm looking to integrate [(-1)/(2-3x)]

is it right to take 1/3 outside the integration symbol to give:

+1/3f[(-1)/(2/3 -x)]
(1/3) ln |(2/3)-x| + c

or should it be:

+1/3f[(-3)/(2-3x)
(1/3) ln |2-3x| +c

The answers gives the bottom one but I got the top one - am I wrong? why?

Thanks

Originally Posted by Bladeowen

Hi everyone - very quick one here

Ill use f in bold do denote integration symbol because I havn't fully mastered the thing to enter formulae yet

Anyway I'm looking to integrate [(-1)/(2-3x)]

is it right to take 1/3 outside the integration symbol to give:



or should it be:


The answers gives the bottom one but I got the top one - am I wrong? why?

Thanks

Are you trying to integrate [-1/(2-3x)] dx?

If so, you CANNOT take 1/3 outside the integration symbol.

Use Substution here:

Suppose that u = 2-3x
Then, du = -3 dx
so, dx = -du/3

substitute this value of dx in the question and you get,

Integral of [-1/u] -du/3
= Integration of 1/3 [1/u] du
= 1/3* ln(u) +c
substitute u = 2-3x, and you have
1/3 * ln(2-3x) + c

which is your answer...

DO not forget to substitute in this case

Ahh yes - many thanks

Wait sorry can you please show me Integral of

2/(1-x)

In the same way? Sorry its basic stuff...

Originally Posted by Bladeowen

Wait sorry can you please show me Integral of

2/(1-x)

In the same way? Sorry its basic stuff...

A lil bit of thinking and you are done!

Suppose that u = 1-x
Then, du = -dx
so, dx = -du

substitute this value of dx in the question

You will be able to do the rest if you follow the previous answer!

You should get:

-2*ln(1-x)+C

Oh sorry my mind was fried there - I'm an idiot

I got dx=-du but forgot I was Integrating wrt u and not x so I put (1-x) back into where u was before I'd integrated!

Idiot lol

Thanks again