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I have this integral to evaluate:

$$\int{\frac{x^4 + 1}{x^2 +1}}\, dx$$

I have tried substitution, trig identity and integration by parts but I'm just going round in circles.

Can anyone explain the method I need to work this out?

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HINT:

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

$$\frac{x^4 + 1}{x^2 +1}=\frac{x^4-1+2}{x^2 +1}=x^2-1+\frac2{x^2+1}$$

Generalization :

If the integrand is $$\frac{a_0+a_1x+a_2x^2+a_3x^3+\cdots}{ax^2+bx+c},$$

just divide to get the quotient of the form $b_0+b_1x+b_2x^2+\cdots$ which is easily integrable and the remaining part will be of the form $\frac{px+q}{ax^2+bx+c}$

If $p=0$ check the $I_2$ below.

else we set $px+q=r\frac{d(ax^2+bx+c)}{dx}+ s=r(2ax)+rb+s$

Comparing the coefficients of $x,2ar=p\implies r=\frac p{2a}$

and Comparing the constants, $rb+s=q\implies s=q-rb=q-\frac{pb}{2a}$

$$\text{So, }\int \frac{px+q}{ax^2+bx+c} dx$$

$$=r \int \frac{d(ax^2+bx+c)}{dx}\frac1{ax^2+bx+c}dx+s\int\frac1{ax^2+bx+c}dx $$

$$=\frac p{2a}\int \frac{d(ax^2+bx+c)} {ax^2+bx+c}+\left(q-\frac{pb}{2a}\right)\int\frac1{ax^2+bx+c}dx $$

$$\text{Now, } \int \frac{d(ax^2+bx+c)} {ax^2+bx+c}=\ln|ax^2+bx+c|+C$$

$$\text{and } I_2=\int\frac1{ax^2+bx+c}dx =\int\frac{4a}{(2ax+b)^2+4ac-b^2}dx $$

If $4ac-b^2=0,$ put $2ax+b=u$ in $I_2$

If $4ac-b^2>0, 4ac-b^2=t^2$(say, ) $I_2=4a\int \frac{dx}{(2ax+b)^2+t^2}$ and put $2ax+u=t\tan\theta$

If $4ac-b^2<0, 4ac-b^2=-t^2$(say, ) $I_2=4a\int \frac{dx}{(2ax+b)^2-t^2}$ and put $2ax+u=t\sec\theta$

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  • $\begingroup$ Why isn't this the best answer? $\endgroup$ – Piano Land Aug 22 '18 at 14:28
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Whenever you have a rational polynomial with a numerator of equal or larger degree than the denominator, try to factor the numerator if possible, or simply, use polynomial long division.

  • lab bhattacharjee noticed a nice way to simplify the rational integrand by manipulating the numerator to make life easier.
  • But suppose you're tired and not feeling particularly creative; polynomial long division will give you the same result:

Dividing numerator by denominator using simple but quick polynomial long division gives us:

$$\frac{x^4 + 1}{x^2 +1}=(x^2-1)+\frac2{x^2+1}$$

So, we can express our integral as follows: $$\int \frac{x^4 + 1}{x^2 +1} \, dx\;= \;\;\int x^2 \,dx \;\;- \;\;\int \,dx \;\;+\;\; 2\int \frac 1{x^2+1}\,dx$$

No doubt, you can handle the first two integrals. And for the third: notice that the third integral is in perfect form which integrates nicely to $$2\int \frac 1{x^2+1}\,dx\;\; = \;\;\;2\tan^{-1}(x) + C$$

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  • $\begingroup$ Deserves another thumbs - up! +1 I have been having a nuts day with a flat tire and all - so am running around like a mad man! :-) Hope your day is better!!! $\endgroup$ – Amzoti May 21 '13 at 0:28
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I think lab bhattacharjee's way is probably the best, but since you mentioned you tried it, I believe trig substitution would work as well. The substitution

$$x=\tan t,dx=\sec^2tdt$$

would result with

$$\int\tan^4t+1=\int(\sec^2t-1)\tan^2t+1dt$$ $$\int\tan^2t\sec^2tdt+\int1-(\sec^2t-1)dt$$

Clear the parentheses, integrate, and backsubstitution should be rather easy.

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Let's try some imaginary stuff.

$\dfrac{x^4}{x^2+1}=\dfrac{Ax^3}{x+i}+\dfrac{Bx^3}{x-i} \implies A=B= \dfrac{1}{2}$

$\dfrac{x^4+1}{x^2+1}= \dfrac{1x^3}{2(x+i)}+\dfrac{1x^3}{2(x-i)}+\dfrac{1}{x^2+1}$

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