# Integral of $\int \frac{x^4+2x+4}{x^4-1}dx$ [duplicate]

I am trying to solve this integral and I need your suggestions. $$\int \frac{x^4+2x+4}{x^4-1}dx$$ Thanks

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## marked as duplicate by Américo Tavares, Lord_Farin, Inceptio, O.L., azimutMay 14 '13 at 15:34

What have you tried? You could use partial fraction decomposition to reduce the integral to table integral. –  Sasha May 14 '13 at 14:55

use polynomial division, we get $$\int \frac{x^4+2x+4}{x^4-1} dx = \int 1 + \frac{2x+5}{(x^2 - 1)(x^2 + 1)}dx = \int 1 + \frac{2x+5}{(x+1)(x-1)(x^2+1)} dx$$

Expressing this as partial fractions, we need only find $A, B, C$

$$= \int \left(1 + \frac{A}{x+1} + \frac B{x-1} +\frac{CX+D}{x^2 + 1}\right)\,dx$$

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If you know where it has been answered, it is good to post the link in the comments to the question. If you put some text in square brackets and follow with the link in parentheses, the text will print and be a link. –  Ross Millikan May 14 '13 at 15:01

Note that this problems cries out for partial fraction decomposition, for which you'll want to factor you denominator: it factors nicely as the difference of squares:

$$(x^4 - 1) = (x^2 + 1)(x^2 - 1) = (x^2 + 1)(x +1)(x-1)$$

See also Paul's Online Notes: Partial Fractions for more insight into using this technique for integration, and some nicely worked out examples.

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Nice guidance +1 –  Amzoti May 15 '13 at 0:31

HINT:

Using Partial Fraction Decomposition formula,

$$\frac{x^4+2x+4}{x^4-1}=1+\frac{ax+b}{x^2+1}+\frac c{x+1}+\frac d{x-1}$$ where $a,b,c,d$ are arbitrary constants to determined by equating the coefficients of the different powers of $x$ in

$$x^4+2x+4=x^4-1+(x^2-1)(ax+b)+c(x-1)(x^2+1)+d(x+1)(x^2+1)$$

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Great formula you shared here! –  NasuSama May 14 '13 at 14:58
@NasuSama, please have a look into the link –  lab bhattacharjee May 14 '13 at 15:01
@lab_bhattarchjee, Thanks for the link. –  NasuSama May 14 '13 at 15:02

Welcome to Math Stack Exchange!

With lab bhattacharjee's helpful hint for the integral, we can rewrite the expression as:

$$\int \left(1 + \frac{(-2x - 5)}{2(x^2 + 1)} + \frac{7}{4(x - 1)} - \frac{3}{4(x + 1)}\right)dx$$

$$= \int 1 dx + \int \frac{(-2x - 5)}{2(x^2 + 1)} dx + \int \frac{7}{4(x - 1)} dx - \int \frac{3}{4(x + 1)} dx$$

$$= \int 1 dx - \int \frac{2x}{2(x^2 + 1)} dx - \int \frac{5}{2(x^2 + 1)} + \int \frac{7}{4(x - 1)} dx - \int \frac{3}{4(x + 1)} dx$$

$$= \int 1 dx - \int \frac{x}{x^2 + 1} dx - \frac{5}{2} \int \frac{1}{x^2 + 1} dx + \frac{7}{4} \int \frac{1}{x - 1} dx - \frac{3}{4} \int \frac{1}{x + 1} dx$$

Next steps should be "direct-cut", meaning that it's really possible to work out the computation one way! :D

• For the first integrand, use power rule to obtain $x$.
• For the second integrand, let $u = x^2 + 1 \rightarrow du = 2x dx$.
• For the third integrand, you get $\arctan(x)$. You can find the special integral here.
• For the fourth integrand, let $v = x - 1 \rightarrow dv = dx$.
• For the fifth integrand, let $w = x + 1 \rightarrow dw = dx$.

So we are left off with:

$$x - \frac{5}{2}arctan(x) - \frac{1}{2} \int \frac{du}{u} + \frac{7}{4} \int \frac{dv}{v} - \frac{3}{4} \int \frac{dw}{w}$$

Finally, the expression becomes...

$$x - \frac{5}{2}arctan(x) - \frac{ln(u)}{2} + \frac{7ln(v)}{4} - \frac{3ln(w)}{4} + K \text{ where } K \text{ is an arbitrary constant}$$ $$= x - \frac{5}{2}arctan(x) - \frac{ln(x^2 + 1)}{2} + \frac{7ln(x - 1)}{4} - \frac{3ln(x + 1)}{4} + K$$

The solution for the integral is equivalent to the one in Wolfram Alpha.

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