How do I integrate 2/(1+x^4) to find the area between two curves?

[liquidheineken]

Find the Area between the two functions.

http://www4a.wolframalpha.com/Calculate/MSP/MSP16151chba72gif4040fg0000686h2hea7f943994?MSPStoreType=image/gif&s=54&w=381.&h=306.&cdf=Coordinates&cdf=Tooltips

I know the bounds are from x=[-1,1] which gives me the equation...
∫ 2/(1+x⁴) - x² dx

I just can't figure out how to integrate 2/(1+x⁴). Substitution won't work.

 

[mooncrater]

This type of integration is a special case.
To do it:
Do you see that 2 in the numerator?
1)Convert that into $1+x^2+1-x^2$.
2)separate The two fractions... I. E.
$\frac {1-x^2}{1+x^4}$+$\frac {1+x^2}{1+x^4}$
3)devide numerator and denominator by $x^2$.
4) know that the differential of $x+\frac {1}{x}$ is $1-\frac {1}{x^2}$.
I think these hints are sufficient.

 

[Mark44]

Find the Area between the two functions.

http://www4a.wolframalpha.com/Calculate/MSP/MSP16151chba72gif4040fg0000686h2hea7f943994?MSPStoreType=image/gif&s=54&w=381.&h=306.&cdf=Coordinates&cdf=Tooltips

I know the bounds are from x=[-1,1] which gives me the equation...
∫ 2/(1+x⁴) - x² dx

Minor point, but the above is NOT an equation -- there's no =, which is the hallmark of an equation.

I just can't figure out how to integrate 2/(1+x⁴). Substitution won't work.

I would be inclined to try a trig substitution.

BTW, when you post a problem, do not delete the homework template. The three parts are required.

 

[Mark44]

This type of integration is a special case.
To do it:
Do you see that 2 in the numerator?
1)Convert that into $1+x^2+1-x^2$.
2)separate The two fractions... I. E.
$\frac {1-x^2}{1+x^4}$+$\frac {1+x^2}{1+x^4}$
3)devide numerator and denominator by $x^2$.
4) know that the differential of $x+\frac {1}{x}$ is $1-\frac {1}{x^2}$.

This is true, but I don't see how it is relevant to this problem. If you divide numerator and denominator of step 2 by x², you get $\frac{1/x^2 - 1}{1/x^2 + x^2} + \frac{1/x^2 + 1}{1/x^2 + x^2}$. I don't see how this is an improvement.

I think these hints are sufficient.

 

[phion]

I would also try a trigonometric substitution.

 

[mooncrater]

This is true, but I don't see how it is relevant to this problem. If you divide numerator and denominator of step 2 by x², you get $\frac{1/x^2 - 1}{1/x^2 + x^2} + \frac{1/x^2 + 1}{1/x^2 + x^2}$. I don't see how this is an improvement.

You can write $x^2+1/x^2$ as $[x+1/x]^2-2$ as well as $[x-1/x]^2+2$. There fore when you take $t=x+1/x$ you get $dt=1-1/x^2$ and when you take $n=x-1/x$ then you get $dn=1+1/x^2$...
These substitutions can be used to calculate the integral.
Well.. I also would like to see the trigonometric substitution method.
(Are you talking about substituting $x^2=tan\theta$? )

 

[phion]

I just plugged the integral into Mathematica and it appears to be kind of a nightmare to solve.

 

[mooncrater]

I just plugged the integral into Mathematica and it appears to be kind of a nightmare to solve.

Mathematica? What does it do?

 

[phion]

 

[mooncrater]

Nightmare! I agree.

 

[mooncrater]

Some integrals have special(and shorter) methods. That just needs recognising the integral type. Otherwise they can be nightmares!