Help with this integral (1 Viewer)

[nancylime]

View attachment Untitled.zip

I can only save it as a zip file sorry!
thanks for the help xx

 

[dan964]

You will need to split it
into thus:
0.5*(2x-3)/(x^2 -3x+4) +1.5/(x^2-3x+4)

the first is a log, the second is a inverse tan.
You should be able to do the rest.

 

[nancylime]

You will need to split it
into thus:
0.5*(2x-3)/(x^2 -3x+4) +1.5/(x^2-3x+4)

the first is a log, the second is a inverse tan.
You should be able to do the rest.

How do you split it into that when the numerator is only x?

 

[D94]

How do you split it into that when the numerator is only x?

Just like how you can split the number 10 into 2 parts, i.e. 5+5 or 7+3, or say, 15-5.

You can split x into 2 parts as well, such as x/2 + x/2 or 2x/3 + x/3 or 2x-x.

So you can even go beyond something like this and do say (x+3)-3 or (x-5)+5, or 0.5(2x-3)+ 3/2.


Expand it out and see.

 

[kawaiipotato]

How do you split it into that when the numerator is only x?

The rule is to make the top look like the derivative of the bottom.