So Can you find the integral of \frac{3x^2-4x+5}{(x-1)(x^2+1)} \ dx?

[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

 

[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.

 

[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 I am too lazy to double check but i think the general idea is enough for u to be able to do it.

Sorry i don't know how to draw the integral sign or fractions otherwise it could have been expressed with more clarity.

 

[CompuChip]

(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

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 ;)

 

[MrSparky]

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 ;)

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

 

[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?

Thanks in advance

 

[CompuChip]

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

 

[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.

 

[optics.tech]

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

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!

 

[The Dagda]

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!

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

 

[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

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

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

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

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