Question

integrate ln(x^2-x+2)

Answer 1

\[\int\limits_{?}^{?}\ln (x^{2}-x+2)dx\]

Answer 2

integral ln(2-x+x^2) dx
Integrate by parts, integral f dg = f g- integral g df,
where
f = ln(x^2-x+2), dg = dx,
df = (2 x-1)/(x^2-x+2) dx, g = x:
= x ln(x^2-x+2)- integral (x (2 x-1))/(x^2-x+2) dx
For the integrand (x (2 x-1))/(x^2-x+2), do long division:
= x ln(x^2-x+2)- integral ((x-4)/(x^2-x+2)+2) dx
Integrate the sum term by term:
= x ln(x^2-x+2)- integral (x-4)/(x^2-x+2) dx- integral 2 dx
Rewrite the integrand (x-4)/(x^2-x+2) as (2 x-1)/(2 (x^2-x+2))-7/(2 (x^2-x+2)):
= x ln(x^2-x+2)- integral ((2 x-1)/(2 (x^2-x+2))-7/(2 (x^2-x+2))) dx- integral 2 dx
Integrate the sum term by term and factor out constants:
= x ln(x^2-x+2)+7/2 integral 1/(x^2-x+2) dx-1/2 integral (2 x-1)/(x^2-x+2) dx- integral 2 dx
For the integrand (2 x-1)/(x^2-x+2), substitute u = x^2-x+2 and du = 2 x-1 dx:
= -1/2 integral 1/u du+x ln(x^2-x+2)+7/2 integral 1/(x^2-x+2) dx- integral 2 dx
For the integrand 1/(x^2-x+2), complete the square:
= -1/2 integral 1/u du+x ln(x^2-x+2)- integral 2 dx+7/2 integral 1/((x-1/2)^2+7/4) dx
For the integrand 1/((x-1/2)^2+7/4), substitute s = x-1/2 and ds = dx:
= 7/2 integral 1/(s^2+7/4) ds-1/2 integral 1/u du+x ln(x^2-x+2)- integral 2 dx
The integral of 1/(s^2+7/4) is (2 tan^(-1)((2 s)/sqrt(7)))/sqrt(7):
= sqrt(7) tan^(-1)((2 s)/sqrt(7))-1/2 integral 1/u du+x ln(x^2-x+2)- integral 2 dx
The integral of 1/u is ln(u):
= sqrt(7) tan^(-1)((2 s)/sqrt(7))-(ln(u))/2+x ln(x^2-x+2)- integral 2 dx
The integral of 2 is 2 x:
= sqrt(7) tan^(-1)((2 s)/sqrt(7))-(ln(u))/2+x ln(x^2-x+2)-2 x+constant
Substitute back for s = x-1/2:
= -(ln(u))/2+x ln(x^2-x+2)-2 x+sqrt(7) tan^(-1)((2 x-1)/sqrt(7))+constant
Substitute back for u = x^2-x+2:
= x ln(x^2-x+2)-1/2 ln(x^2-x+2)-2 x+sqrt(7) tan^(-1)((2 x-1)/sqrt(7))+constant
Which is equal to:
= (x-1/2) ln(x^2-x+2)-2 x+sqrt(7) tan^(-1)((2 x-1)/sqrt(7))+constant

Answer 3

@phystmg ok, now can you do that using the "SIGMA Equation" function?

Answer 4

If you want to do a definite integral with \(\sum\) notation on an interval \([a, b]\), it would look like\[\left(b-a \over n\right)\sum_0^{n-1} {\ln(\left(a+{b-a \over n}\right)^2-\left(a+{b-a \over n}\right)+1) }\]Where a greater \(n\) corresponds to greater accuracy.

Answer 5

Sorry, when I said sigma, I meant that blue button that says "Equation" and has a sigma.

Answer 6

I can't translate all that wording. It's confusing.

Answer 7

Oh well. It looks to me like phystmg just copy pasted http://www.wolframalpha.com/input/?i=ln%28x%5E2-x%2B2%29

Scroll down to where it comes up with the indefinite integral and click on show steps. It's exactly what phystmg did.

Answer 8

You've got to be kidding...although useful that's just lame. If you're going to explain something, you should at least know what you're explaining.

Answer 9

Sometimes it's just too much of a pain to write out everything yourself. And btw, I had the wrong link. It's actually http://www.wolframalpha.com/input/?i=ln%28x%5E2-x%2B1%29

The first one had \(x^2-x+2\) instead \(x^2-x+1\)

Answer 10

Don't worry, you got it right the first time. It was a 2, not 1. I do appreciate the link though.

Answer 11

I'm getting tired :(

I'm not sure if there's a faster/easier way to do this. WolframAlpha generally does integrals in the worst way possible.

Answer 12

lol