indefinite integral

**Random Variable** · December 4th, 2010, 08:20 AM

Challenge Problem:

$\int \frac{x^{2}-1}{(x^{2}+1)\sqrt{x^4+1}} \ dx$

Moderator approved CB

**Random Variable** · December 4th, 2010, 10:33 AM

My solution is only valid for certain intervals.

**TwistedOne151** · December 4th, 2010, 07:49 PM

For the interval x>0, we can use the u-substitution
$u=\frac{x^2-1}{x^2+1}$ , so that
$x^2=\frac{1+u}{1-u}$ ,
$x=\frac{\sqrt{1+u}}{\sqrt{1-u}}$ ,
$dx=\frac{du}{\sqrt{1+u}(1-u)^{3/2}}=\frac{du}{(1-u)\sqrt{1-u^2}}$ .
Thus,
$\int\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}\,dx=\int\frac{u}{\sqrt{\le ft(\frac{1+u}{1-u}\right)^2+1}}\frac{du}{(1-u)\sqrt{1-u^2}}$
$\;\;=\int\frac{u(1-u)}{\sqrt{(1+u)^2+(1-u)^2}}\frac{du}{(1-u)\sqrt{1-u^2}}$
$\;\;=\int\frac{u}{\sqrt{2+2u^2}\sqrt{1-u^2}}\,du$
$\;\;=\frac{1}{\sqrt2}\int\frac{u}{\sqrt{1-u^4}}\,du$
$\;\;=\frac{1}{2\sqrt2}\int\frac{2u}{\sqrt{1-u^4}}\,du$
$\;\;=\frac{1}{2\sqrt2}\sin^{-1}u^2+C$
$\;\;=\frac{1}{2\sqrt2}\sin^{-1}\left(\frac{x^2-1}{x^2+1}\right)^2+C$

--Kevin C.

**Random Variable** · December 5th, 2010, 08:02 AM

My solution looks different, but it only differs by a constant.

$\int \frac{x^{2}-1}{(x^{2}+1)\sqrt{x^4+1}} \ dx$

$= \int \frac{x^{2}-1}{(x^{2}+1)\sqrt{x^4+1}} \ dx \ \frac{1/x}{1/x} \ \text{if} \ x \ne 0$

$= \int \frac{(x - 1/x)}{x(x+ 1/x) \sqrt{x^{2}+1/x^{2}}} \ dx \ \text{if} \ x>0$

$= \int \frac{(x - 1/x)}{x (x+ 1/x) \sqrt{(x+1/x)^{2} -2}} \ dx$

$\text{let} \ x =e^{u}$

$=\int \ \frac{(e^{u} - e^{-u})}{(e^{u}+ e^{-u}) \sqrt{(e^{u}+e^{-u})^{2} -2}} \ du$

$\text{let}\ v=e^{u}+e^{-u}$

$= \int \frac{1}{v \sqrt{v^{2} -2}} \ dv$

$\text{let} \ v = \sqrt{2} \sec w$

$= \frac{1}{\sqrt{2}} w + D$

$= \frac{1}{\sqrt{2}} \sec^{-1} \Big(\frac{x + 1/x}{\sqrt{2}} \Big) + D$

where $D = C + \frac{\sqrt{2} \pi}{8}}$

**Random Variable** · December 5th, 2010, 08:17 AM

But the function we're integrating is even, so for x<0, $\int \frac{x^{2}-1}{(x^{2}+1)\sqrt{x^4+1}} \ dx = - \frac{1}{\sqrt{2}} \sec^{-1} \Big(\frac{|x| + 1/|x|}{\sqrt{2}} \Big) + D$ , no?

**Krizalid** · January 26th, 2011, 12:33 PM

hadn't even seen this section about challenge problems, it's pretty interesting.

perhaps this is one of the hardest integral i've ever seen, but with a little of magic we can take it down easily.

note that

$\displaystyle\int{\frac{{{x}^{2}}-1}{\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{4}}+1}}\,dx}=\int{\frac{1-\frac{1}{{{x}^{2}}}}{\left( x+\frac{1}{x} \right)\sqrt{{{\left( x+\frac{1}{x} \right)}^{2}}-2}}\,dx},$

whereat the substitution $x+\dfrac1x=\dfrac{\sqrt2}t$ will solve the problem.

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