what is the integral of (x^2+x-1) / (x^2-x) dx? pls help none of my multiple choice answers match wat i got?

Answer 1

Integral [ (x^2 + x - 1) / (x^2 - x) ] dx

Rule of thumb: Whenever the degree of the numerator is greater than or equal to the degree of the denominator, you should use synthetic long division.

Without showing you the details, you should get that

(x^2 + x - 1) / (x^2 - x) = {(2x - 1) / (x^2 - x)} + 1

So our new integral now becomes

Integral [ {(2x - 1) / (x^2 - x)} + 1 ] dx

We can separate this into two integrals.

Integral [ (2x - 1) / (x^2 - x)] dx + Integral (1) dx

Integral [ (2x - 1) / (x^2 - x)] dx + x

At this point, it is worth noting that if we use u substitution and let u = x^2 - x, du = (2x - 1) dx. This becomes EXTREMELY useful, and I'll show you why by rearranging the integral itself.

Integral [ 1/(x^2 - x) * (2x - 1) dx ] + x

Note that we can replace (2x - 1) ENTIRELY with du, getting us

Integral [ (1/u) * du ] + x

And now, we can solve the integral normally.

ln|u| + x + C

Putting u = x^2 - x back,

ln |x^2 - x| + x + C

In theory, with great algebraic difficulty, this can be combined into one term, which MAY explain why none of the multiple choice answers works.

x can be written as ln(e^x), and c can be written as ln(e^c), which is a constant anyway which we'll call k. Thus, your answer might be:

ln | k (e^x) (x^2 - x) |

Let's hope that's not the case though.

Answer 2

I get ln(x) + 2x + C

Rearrange the top to x^2 - 1 + x
x^2 - 1 can be factored to (x + 1)(x - 1), so the top becomes (x + 1)(x - 1) +x

Factor out x from the bottom and get x(x - 1)

Divide top and bottom by (x - 1) and the top becomes (x + 1) + x, which simplifies to 2x + 1.

The bottom becomes x.

Separate the top and you get the ingegral of (1/x + 2)dx.
That evaluates to ln(x) + 2x + C.

Answer 3

(x^2+x-1)/(x^2-x) = {(x^2-x) +x + (x-1)}/(x^2-x)
= 1 + 1/(x-1) + 1/x
So, the integral of it is:
I = x + ln(x-1) + ln(x) + C, where C is a constant.

Answer 4

As y=(xx+x-1) /(xx-x) = 1 + (2x-1)/x/(x-1) = 1 + a/x +b/(x-1);
Or a/x +b/(x-1) = {(a+b)x –b} / x/(x-1), thence a+b=2, -b=-1, hence a=b=1;
Return to y= 1+ 1/x +1/(x-1); thus I= C +x +ln|xx-x|;

Answer 5

(x^2+x-1) / (x^2-x)
=1 + 1/(x-1) + 1/x

Therefore,

∫(x^2+x-1) / (x^2-x)
= ∫1 + 1/(x-1) + 1/x
= x + ln|x-1| + ln|x| + c

Answer 6

first find the derivative
so you can rewrite it as (x^2+x-1)(x^2-x)^-1 so now find it's derivative

Answer 7

i havent got that far into math yet. i am taking my GED i am in the integers part.

Answer 8

(x²+x-1) / (x²-x) dx
(x+1)(x-1) / x(x-1) dx
x+1 / x
::

Answer 9

most of the time things get really confusing so you could be right and the answers wrong