Complicated Integral evaluation

Question

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!

Answer 1

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

Answer 2

hint

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

Answer 3

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 $u=\sqrt2 \tan\theta\to\text{d}u=\sqrt2\sec^2 \theta\text{d}\theta\implies \int\frac{\sqrt2\sec^2\theta}{\sqrt2\tan\theta\sqrt2\sec\theta}\text{d}\theta\to\frac{1}{\sqrt2}\int\csc\theta\text{d}\theta=$ $\frac{1}{\sqrt2}\ln \left (\csc\theta-\cot\theta\right )=\frac{1}{\sqrt2}\ln\left (\frac{\sqrt{u^2+2}-\sqrt2}{u}\right )=$ $\frac{1}{\sqrt2}\ln\left (\frac{\sqrt{x^4+1}-x\sqrt2}{x^2-1}\right )+C$