Question

integral (x+2)/(x^2 -x +1) , would i use integration by parts?

Answer 1

id go with the downside of x+2 and the upside being ... well, the other

Answer 2

but thats just a gut thing

Answer 3

trial and error; thats just how i roll :)

Answer 4

okay and then how would i do x/x^2 +x +1

Answer 5

Actually you should break it to \(\frac{x-1}{x^2-x+1}+\frac{3}{x^2-x+1}\).

Answer 6

No, that wouldn't work either!

Answer 7

where'd you get the 3?

Answer 8

Put it as \(\frac{x-\frac{1}{2}}{x^2-x+1}+\frac{\frac{5}{2}}{x^2-x+1}\). I think this will work.

Answer 9

Better, sorry, write

\[\frac {x+2}{x^2 -x +1} = \frac{(1/2)(2x - 1)}{x^2 - x + 1} + \frac{5/2}{x^2 - x + 1}\]

Answer 10

why are you making it 1/2 and 5/2?

Answer 11

i dont understand the numerators in either fraction

Answer 12

Yes. We wrote it that way because you now use an easy substitution in the first term:

u = x^2 - x + 1

Answer 13

For the first term, use a substitution \(u=x^2-x+1\). That gives \(\frac{1}{2}\int \frac{1}{u}du=\frac{1}{2}\ln |u|=\frac{1}{2}\ln|x^2-x+1|\).

Answer 14

The second term is a bit more complicated, you need to complete the square in the bottom.

Answer 15

okay so it is 1/2ln(x^2-x+1) + 5/2ln(x^2 -x+1)?

Answer 16

....yes, as we talked about before

\[x^2 - x + 1 = x^2 - x + 1/4 + 3/4 = (x-1/2)^2 + (\sqrt{3}/2)^2 \]

Answer 17

Then you use a tan substitution.

I.e., how would you evaluate:\[\int\limits \frac{1}{1 + t^2} \ dt\]

Answer 18

is my answer right though?

Answer 19

No, it's not. The second term is NOT an expression in log

Answer 20

(5/2)/(x^2 -x +1) = 5/2ln(x^2 -x+1) isn't true?

Answer 21

ohhhh no it isn't!

Answer 22

okay i get it now, how would i do it though? sorry i know you just expained it

Answer 23

The result of the second term would be arctan(something).

Answer 24

why do i need o complete the square in the bottom?

Answer 25

I'll tell you so you can work towards it:
\[\frac{5}{2} \int\limits \frac{1}{x^2 - x + 1} \ dx = \frac{5}{\sqrt{3}} \tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)\]

Answer 26

how did you get that though?

Answer 27

Write
\[\frac{1}{x^2 - x +1} = \frac{1}{(x-1/2)^2 + (\sqrt{3}/2)^2}\]

Answer 28

Now substitute x - 1/2 = sqrt(3)/2. tan u

Answer 29

ohhh okay i got it thanks!

Answer 30

\[\int\limits_{}^{}{{5 \over 2} \over (x-{1 \over 2})^2+{3 \over 4}}dx=\int\limits_{}^{}{{5 \over 2} \over {3 \over 4}({2 \over \sqrt{3}}(x-{1 \over 2})^2+1}={10 \over 3}\int\limits_{}^{}{1 \over ({2x-1 \over \sqrt{3}})^2+1}dx\]
Now substitute \(u=\frac{2x-1}{\sqrt{3}} \implies du=\frac{2}{\sqrt{3}}dx\). The integral then becomes:
\({5 \over \sqrt3}\int\limits_{}^{}{1 \over u^2+1}={5 \over \sqrt3} \tan^{-1}u+c\). Just substitute back to get your answer.

Answer 31

I think I did almost all the steps you need. There are a couple of intermediate steps that only include simple algebra, like taking a common factor or opening a parenthesis.