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Here is an integral I have been struggling with:

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

Here is what I have done so far:

First, I make some cancellations

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

Notice that because $(1+x^2)-(1+x^2) = 0$, the top and the bottom $1+x^2$ will cancel, and then we can factor the bottom

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

Now I make some simplifications to the equation

$$\int{\frac{dx}{1-x^2}} = \int{\frac{1}{(1-x)}dx\int{\frac{1}{(1+x)}}}dx$$ $$d\int{\frac{x}{1-x^2}} = (-\ln(x)+1)(\ln(x)+1)$$ $$d\int{\frac{x}{1-x^2}} = 1-\ln^2(x)$$ $$d\int{(\frac{x}{1}-\frac{x}{x^2})} = 1-1n^2(x)$$ $$d\int{(x-\frac{1}{x})} = 1-n^2(x)$$ $$d(\frac{1}{2}x^2-\ln(x)) = 1-n^2(x)$$ $$d = \frac{1-n^2(x)}{\frac{1}{2}x^2-\ln(x)}$$ So I solved for $d$, but this isn't one of the multiple choice options...

Any ideas?

Thanks!

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    $\sqrt{1+x^4}$ is not at all the same as $\sqrt{(1+x^2)^2}.\qquad$ – 2017-01-10
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    factoring $d$ out of the integral sign seriously makes my day. I have never seen this before in 25+ years of being involved in high level math :) This is even better than $$\frac{\sin x}{n} = \text{six} = 6$$ – 2017-01-10
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    BTW: http://www.wolframalpha.com/input/?i=integrate%5B(1-x%5E2)%2F(Sqrt%5B1%2Bx%5E4%5D(1%2Bx%5E2)),x%5D – 2017-01-10
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    @gt6989b But everything else seems good, right? I don't see any other errors. – 2017-01-10
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    @Tdonut this is epic enough :) but there is more -- how about $$\int f \times g = \int f \times \int g?$$ – 2017-01-10
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    @gt6989b Yeah that's an error, but other than that? – 2017-01-10
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    @Tdonut how about $1+x^4 = \left(1+x^2\right)^2$? – 2017-01-10
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    @gt6989b yeah yeah yeah but more than that? – 2017-01-10
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    @Tdonut that's enough for one question :) This quite likely already sets a Math.SE record :) – 2017-01-10
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    @gt6989b I'm confused about what $n$ is equal to. Is it any integer? – 2017-01-10
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    Voting to close as too broad (zing!). Plus, I don't really think it's been asked in good faith. – 2017-01-10
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    This is my new favorite question, I like how $\ln^2(x) = 1 \cdot n^2(x) = n^2(x)$ was done – 2017-01-10
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    @gt6989b The same thing, in approx. a double time. – 2017-01-10
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    But wait... there is more :) how about integral transform in the last couple steps, apparently $$\frac{x}{1-x^2} = \frac{x}{1} - \frac{x}{x^2} = x - \frac{1}{x}$$ – 2017-01-10

3 Answers 3

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Your first issue was assuming $$ 1+x^4 = (1+x^2)^2 $$ (which it does not)

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    Hmm... That's a gray area, but were my steps after that correct? – 2017-01-10
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    Those are the only two mistakes I see, though. Everything else seems good. – 2017-01-10
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    @Tdonut How about on the left side $$\int \frac{dx}{...} = d \int \frac{x}{...}$$? – 2017-01-10
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    Or how $\ln$ became $n$ – 2017-01-10
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hint

the substitution $x^2=\sinh(t)$ should work.

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Divide both top and bottom by $x^2\implies\int \frac{1+\frac{1}{x^2}}{\left (x-\frac{1}{x}\right )\sqrt{x^2+\frac{1}{x^2}}}\text{d}x$ conveniently $x^2+\frac{1}{x^2}=\left (x-\frac{1}{x}\right )^2+2$ so making the substitution $u=x-\frac{1}{x}\to\text{d}u=\left (1+\frac{1}{x^2}\right )\text{d}x\implies\int\frac{1}{u\sqrt{u^2+2}}\text{d}u$ making the substitution $u=\sqrt2 \tan\theta\to\text{d}u=\sqrt2\sec^2 \theta\text{d}\theta\implies \int\frac{\sqrt2\sec^2\theta}{\sqrt2\tan\theta\sqrt2\sec\theta}\text{d}\theta\to\frac{1}{\sqrt2}\int\csc\theta\text{d}\theta=$ $\frac{1}{\sqrt2}\ln \left (\csc\theta-\cot\theta\right )=\frac{1}{\sqrt2}\ln\left (\frac{\sqrt{u^2+2}-\sqrt2}{u}\right )=$ $\frac{1}{\sqrt2}\ln\left (\frac{\sqrt{x^4+1}-x\sqrt2}{x^2-1}\right )+C$