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

• jessjolt
In summary, 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/8so plugging in -pi/8 for B makes:x= tan[ (B-(pi/8))x^2 + (B+C
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

Hi jessjolt!

Firstly, you should know that

$$\int{\frac{arctan(x)}{(x+1)(x^2+1)}dx}$$

is not solvable using elementary functions.

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

$$arctan x = A(x^2+1) + (Bx+C)(x+1)$$

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

$$\frac{1}{(x+1)(x^2+1)}$$

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

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!

micromass said:
Hi jessjolt!

Firstly, you should know that

$$\int{\frac{arctan(x)}{(x+1)(x^2+1)}dx}$$

is not solvable using elementary functions.

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

$$arctan x = A(x^2+1) + (Bx+C)(x+1)$$

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

$$\frac{1}{(x+1)(x^2+1)}$$

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?

jessjolt said:
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.

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

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

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.

Char. Limit said:
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, $$\frac{1}{(x+1)(x^2+1)}=\frac{1}{2(x^2+1)}-\frac{x}{2(x^2+1)}+\frac{1}{2 (x+1)}$$
The first term will give $\displaystyle \int{\frac{\tan^{-1}(x)}{2(x^2+1)}}\,dx\,,$ 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:
$$\frac{x^{2n}}{(x^2+1)}=x^{2(n-1)}-x^{2(n-2)}+x^{2(n-3)}-\dots-(-1)^nx^2(-1)^n+\frac{(-1)^n}{x^2+1}$$

The third term can he handled similarly.​

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

## 1. What is the purpose of using partial fractions when integrating?

Partial fractions is a method used to break down a rational function into simpler fractions, making it easier to integrate. It allows us to rewrite the function in terms of simpler integrals that are easier to solve.

## 2. How do you determine the partial fractions for a given rational function?

To determine the partial fractions, we first factor the denominator and then set up a system of equations using the coefficients of the factors. We then solve for the unknown coefficients using algebraic manipulation.

## 3. How do you integrate using partial fractions?

After determining the partial fractions, we integrate each term separately. This can be done by using substitution or by using the formula for integrating powers of x. We then combine the resulting integrals to get the final answer.

## 4. Can partial fractions be used for all rational functions?

No, partial fractions can only be used for proper rational functions, which are functions where the degree of the numerator is less than the degree of the denominator. If the function is improper, we must first divide the numerator by the denominator to get a proper rational function before using partial fractions.

## 5. How does the presence of trigonometric functions, such as arctan x, affect the partial fractions?

The presence of trigonometric functions in the original function does not affect the process of finding the partial fractions. However, when integrating, we may need to use trigonometric identities or substitution to simplify the resulting integrals.

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