Integrate using partial fractions (arctan x) / ((x+1)(x^2+1))?

[jessjolt]

Integrate using partial fractions (arctan x) / ((x+1)(x^2+1))?

My attempt:

(arctan x) / ((x+1)(x^2+1)) = A/(x+1) + (Bx+C)/(x^2+1)
arctan x = A(x^2+1) + (Bx+C)(x+1)
when x= -1, A= -pi/8

so plugging in -pi/8 for B makes:
x= tan[ (B-(pi/8))x^2 + (B+C)x + (C-(pi/8)) ]
and when x=0, C= pi/8

so plugging in pi/8 for C and rearranging produces:
x=tan[ B(x^2+x) + (((pi*x)/8)*(1-x)) ]
and when x=1, B=pi/8

so (arctan x) / ((x+1)(x^2+1)) = [(-pi/8)/(x+1)] + {{ [(pi*x)/8] + (pi/8) }/(x^2+1) }

but when i graph in radian mode on a graphic calculator (arctan x) / ((x+1)(x^2+1)) and my result, they are not the same thing, so my answer must be wrong. The graphs appear very similar though, so it seems there is only a small error? Please help?? :D

 

[micromass]

Hi jessjolt!

Firstly, you should know that

[tex]\int{\frac{arctan(x)}{(x+1)(x^2+1)}dx}[/tex]

is not solvable using elementary functions.

Secondly, what you did is sadly not allowed. You wrote

[tex]arctan x = A(x^2+1) + (Bx+C)(x+1)[/tex]

But this would imply that the arctangent was a polynomial, which is not not the case. What you could do is split

[tex]\frac{1}{(x+1)(x^2+1)}[/tex]

into partial fractions and then multiply it all by arctan(x). But, then again, the integral isn't solvable...

 

[eumyang]

You can't use partial fractions here, because your original function is NOT a rational function. (Rational functions are in the form of f(x)/g(x), where f and g are polynomials.)EDIT: Beaten to it!

 

[jessjolt]

Hi jessjolt!

Firstly, you should know that

[tex]\int{\frac{arctan(x)}{(x+1)(x^2+1)}dx}[/tex]

is not solvable using elementary functions.

Secondly, what you did is sadly not allowed. You wrote

[tex]arctan x = A(x^2+1) + (Bx+C)(x+1)[/tex]

But this would imply that the arctangent was a polynomial, which is not not the case. What you could do is split

[tex]\frac{1}{(x+1)(x^2+1)}[/tex]

into partial fractions and then multiply it all by arctan(x). But, then again, the integral isn't solvable...

oohhh ok i see thanks...so there is no way to use partial fractions with transcendental functions like the arctangent?

 

[micromass]

oohhh ok i see thanks...so there is no way to use partial fractions with transcendental functions like the arctangent?

Not that I'm aware of, no.

 

[SammyS]

Well you can expand [itex]\frac{1}{(x+1)(x^2+1)}[/itex] using partial fractions, but I'm quite that won't help with this integral.

 

[cragar]

you could write the arctan(x) in term of natural logs . and then factor
[itex] (x^2+1) [/itex] as (x+i)(x-i) but i don't know if that will help .

 

[Char. Limit]

I think that if you write tan^-1(x) as an infinite series, you can integrate it then. Of course, it will be hell, and your integral will probably be in a series.

 

[SammyS]

I think that if you write tan^-1(x) as an infinite series, you can integrate it then. Of course, it will be hell, and your integral will probably be in a series.

(OP seems to have dropped out of this discussion.)

Ooooo ... I like this idea ! However, I wouldn't want to work it all out.

But, it does lead to the following hybrid scheme:

Using partial fractions, [tex]\frac{1}{(x+1)(x^2+1)}=\frac{1}{2(x^2+1)}-\frac{x}{2(x^2+1)}+\frac{1}{2 (x+1)}[/tex]
The first term will give [itex]\displaystyle \int{\frac{\tan^{-1}(x)}{2(x^2+1)}}\,dx\,,[/itex] which is not a difficult integration.

For the 2nd & third terms, use the infinite series expansion for arctan(x), expanded about x = 0. (It has only odd powers of x.)

Multiplying this series by the second term results in a series whose terms are of the form:
[tex]\frac{x^{2n}}{(x^2+1)}=x^{2(n-1)}-x^{2(n-2)}+x^{2(n-3)}-\dots-(-1)^nx^2(-1)^n+\frac{(-1)^n}{x^2+1}[/tex]

The third term can he handled similarly.​

Char. Limit is right. This is a mess - but do-able.