Integral question (1 Viewer)

kendricklamarlover101 · Jan 25, 2024

can someone check if my working is correct i got this integral from spivak's calculus
$\int \frac{x^2 -1}{x^2 +1} \frac{1}{\sqrt{1+x^4}} \; dx \quad \text{Using } u^2 = x^2 + \frac{1}{x^2}$
$u^2 = x^2 + \frac{1}{x^2} \implies 2u \; du = \left(2x - \frac{2}{x^3}\right)dx$
$u \; du = \left(x - \frac{1}{x^3}\right)dx$
$= \frac{x^4-1}{x^3} \; dx = \frac{(x^2-1)(x^2+1)}{x^3} \; dx$
$\implies dx = \frac{ux^3 \; du}{(x^2-1)(x^2+1)}$
$\int \frac{x^2 -1}{x^2 +1} \frac{1}{\sqrt{1+x^4}} \; dx = \int \frac{x^2 -1}{x^2 +1} \frac{1}{x\sqrt{x^2 + \frac{1}{x^2}}} \; dx$
$=\int \frac{x^2 - 1}{x^2 + 1} \frac{1}{xu} \frac{ux^3 \; du}{(x^2-1)(x^2+1)} = \int \frac{x^2}{\left(x^2 +1\right)^2} \; du$
$=\int \frac{x^2}{x^2 \left( x +\frac{1}{x} \right)^2} \; du = \int \frac{1}{u^2 + 2} \; du$
$=\frac{1}{\sqrt{2}} \arctan \left( \frac{u}{\sqrt{2}} \right) = \frac{1}{\sqrt{2}} \arctan \left( \frac{\sqrt{x^4+1}}{x\sqrt{2}} \right) + C$
also are there any alternative methods than this substitution?

Drongoski · Jan 25, 2024

I've checked. Seems to be correct. Congratulations.
I would not expect you to have Michael Spivak's textbook.

kendricklamarlover101 · Jan 25, 2024

yea my tutor gave me the textbook cause he thought i would have liked some of the questions. at first i tried using a trig substitution but i couldnt get anywhere with that so i was just wondering if there was another solution.

Drongoski · Jan 25, 2024

kendricklamarlover101 said:
yea my tutor gave me the textbook cause he thought i would have liked some of the questions. at first i tried using a trig substitution but i couldnt get anywhere with that so i was just wondering if there was another solution.

You've done very well to do the integration with the suggested substitution. I tried x^2 = tan@ - fruitless.

tywebb · Jan 25, 2024

kendricklamarlover101 said:
are there any alternative methods than this substitution?

Of course there is:

tywebb · Jan 25, 2024

The one I put there is Michael Spivak's own solution.

It comes from the Combined Answer Book For Calculus Third and Fourth Editions - Chapter 19 Q8v.

The Calculus book and Combined Answer Book are both available as ebooks.

tywebb · Jan 25, 2024

One can improve upon Spivak's solution by combining the 2 substitutions:

$\text{Let }u=\textstyle\left(\frac{x^2-1}{x^2+1}\right)^2.\text{ Then }du=\frac{8x(x^2-1)}{(x^2+1)^3}\ \!dx$

Hence

$\begin{aligned}\textstyle\int\frac{x^2-1}{x^2+1}\cdot\frac{1}{\sqrt{1+x^4}}\ \!dx&=\textstyle\frac{1}{2\sqrt2}\int\frac{x^2-1}{x^2+1}\cdot\frac{8x}{\sqrt{8x^2(1+x^4)}}\ \!dx\\ &=\textstyle\frac{1}{2\sqrt2}\int\frac{(x^2+1)^2}{\sqrt{(x^2+1)^4-(x^2-1)^4}}\cdot\frac{8x(x^2-1)}{(x^2+1)^3}\ \!dx\\ &=\textstyle\frac{1}{2\sqrt2}\int\frac{1}{\sqrt{1-\left(\frac{x^2-1}{x^2+1}\right)^4}}\cdot\frac{8x(x^2-1)}{(x^2+1)^3}\ \!dx\\ &=\textstyle\frac{1}{2\sqrt2}\int\frac{1}{\sqrt{1-u^2}}\ \!du\\ &=\textstyle\frac{1}{2\sqrt2}\sin^{-1}u+C\\ &=\textstyle\frac{1}{2\sqrt2}\sin^{-1}\left(\frac{x^2-1}{x^2+1}\right)^2+C\end{aligned}$

tywebb · Jan 25, 2024

Integrals aside, at 80 Spivak broke his hip and died in 2020.

michael-spivak-072c41ae-14fa-4007-97c5-d29537880f3-resize-750.jpeg

kendricklamarlover101 · Wednesday at 7:52 PM

i recently found another way to do this integral and thought it was pretty cool
$\int \frac{x^2 - 1}{x^2+1}\frac{dx}{\sqrt{1+x^4}} = \int \frac{1-\frac{1}{x^2}}{1+\frac{1}{x^2}}\frac{dx}{\sqrt{x^2\left(x^2+\frac{1}{x^2}\right)}}$
$=\int \frac{ 1 - \frac{1}{x^2} }{\left( x + \frac{1}{x} \right) \left( \sqrt{x^2 + \frac{1}{x^2} } \right)}dx = \int \frac{ 1 - \frac{1}{x^2} }{\left( x + \frac{1}{x} \right) \left( \sqrt{\left(x+\frac{1}{x}\right)^2-2} \right)}dx$
$\text{Let } u=x+\frac{1}{x}, \; du = \left(1 - \frac{1}{x^2}\right)dx$
$I= \int\frac{du}{u\sqrt{u^2-2}}$
$\text{Let } u=\sqrt{2}\sec \theta, \; du=\sqrt{2}\sec\theta\tan\theta d\theta$
$I= \int\frac{\sqrt{2}\sec\theta\tan\theta d\theta }{\sqrt{2}\sec\theta\sqrt{2\sec^2\theta - 2} } =\int\frac{\sqrt{2}\sec\theta\tan\theta d\theta }{\sqrt{2}\sec\theta\sqrt{2}\tan\theta}$
$= \frac{1}{\sqrt{2}}\sec^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{2}}\right) + C$

imagineee · Wednesday at 8:16 PM

Yep that's the way I would've done it, although the trig sub is kind of unnecessary since you can change I to 1/u^2(1-2/u^2)^1/2 then it's essentially reverse chain-ruleable into a standard arcsine/arccos integral

Integral question (1 Viewer)

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