# Integral Question

1. Jan 27, 2009

### optics.tech

Hi,

Can anyone tell me how can I find the integral of $$\frac{3x^2-4x+5}{(x-1)(x^2+1)} \ dx$$ ?

Thanks

2. Jan 27, 2009

### CompuChip

I found a very nice way: what is the derivative of the denominator? Now write the numerator as that + something else and split into two integrations.

3. Jan 28, 2009

### MrSparky

Im not sure if this is the easiest way to do it but u can manipulate the equation to

((x-3)/(x^2 +1))+ (2/x-1)
Then u split it up into a 3rd fraction so what you get is
(x/x^2 +1) -(3/x^2 +1) + (2/x-1) dx
then what u integrate i think u should get
1/2 ln (x^2 +1) - 1/3 ln (x^2 +1) + 2 ln (x-1) + c
I might have done something wrong since im too lazy to double check but i think the general idea is enough for u to be able to do it.

Sorry i dont know how to draw the integral sign or fractions otherwise it could have been expressed with more clarity.

4. Jan 28, 2009

### CompuChip

You're not entirely correct: -1/3 ln(x^2 + 1) gives - (2/3) x / (x^2 + 1) when differentiated, instead of - 3 / (x^2 + 1).

Your approach is probably possible though, although I humbly think that mine is easier ;)

5. Jan 28, 2009

### MrSparky

oops, guess i misread that part when working it out, but yea your method is most likely better than mine.

6. Jan 28, 2009

### optics.tech

Sorry I have forgotten. If I am not wrong, it's a partial fraction:

$$\frac{3x^2-4x+5}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$$

$$=\frac{A(x^2+1)+(Bx+C)(x-1)}{(x-1)(x^2+1)}$$

$$\frac{Ax^2+A+Bx^2-Bx+Cx-C}{(x-1)(x^2+1)}$$

$$\frac{(A+B)x^2-(B-C)x+A-C}{(x-1)(x^2+1)}$$

$$A + B = 3 \ .................... \ (1)$$

$$-(B-C)=-4 \ .................... \ (2)$$

$$A-C=5 \ .................... \ (3)$$

from equation 1, 2, and 3, I can obtain:

$$A=2, B=1, C=-3$$

so the fraction can be modified as:

$$\frac{2}{x-1}+\frac{x-3}{x^2+1}$$

$$=\frac{2}{x-1}+\frac{x}{x^2+1}-\frac{3}{x^2+1}$$

so

$$\int\frac{3x^2-4x+5}{(x-1)(x^2+1)} \ dx=\int(\frac{2}{x-1}+\frac{x}{x^2+1}-\frac{3}{x^2+1}) \ dx$$

$$=2\int\frac{1}{x-1} \ dx+\int\frac{x}{x^2+1} \ dx-3\int\frac{1}{x^2+1} \ dx$$

$$=2 \ ln(x-1)+\frac{1}{2} \ ln(x^2+1)-3 \ arctan \ x+C$$

Is it correct?

7. Jan 28, 2009

### CompuChip

Looks fine to me!
You can always do the differentiation and check that it works out :tongue:

8. Jan 28, 2009

### The Dagda

It is and not only that it was the way I was going to tackle it yesterday if I'd of had time. Kudos.

I've not seen the ABC method before, I'd of done it manually myself. Interesting stuff.

9. Jan 28, 2009

### optics.tech

Hi The Dagda,

You can find the "ABC" method in most of elementary calculus textbook called "partial fraction", I suggest you to search it in the index .

Thanks!

10. Jan 28, 2009

### The Dagda

They didn't teach it that way, but then my course wasn't standard. I have a maths text book I bought separately though, I'll look it up, cheers.

11. Jan 28, 2009

### optics.tech

The last part of above integration $$(\int\frac{3}{1+x^2}dx)$$ is:

The Derivative Rule of Inverse Function:

$$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$$

With

$$y=f(x)=tan \ x$$
$$y^{-1}=f^{-1}(x)=tan^{-1}x$$
$$y'=f'(x)=sec^2x$$

So

$$(f^{-1})'(x)=\frac{1}{sec^2(tan^{-1}x)}$$

since

$$sec^2u=1+tan^2u$$

then

$$(f^{-1})'(x)=\frac{1}{sec^2(tan^{-1}x)}=\frac{1}{1+tan^2(tan^{-1}x)}$$

also

$$tan(tan^{-1}x)=x$$

again

$$(f^{-1})'(x)=\frac{1}{sec^2(tan^{-1}x)}=\frac{1}{1+tan^2(tan^{-1}x)}=\frac{1}{1+x^2}$$

therefore

$$\frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^2}$$

So

$$\int\frac{3}{1+x^2}dx$$

$$=3\int\frac{1}{1+x^2}dx$$

$$=3 \ tan^{-1}x + C$$

$$=3 \ arctan \ x + C$$