Calculus Integration Problem?

Question

Hello, I am currently in calculus 3 and I know this question is from late calculus 2 but I'm definitely rusty and can't figure out the process to solve it. Does anyone know how to solve:

Integral(x/(1-x^8)) ????

Thanks for any help

Answer 1

I did a substitution y = x^2, which gives you a half factor in front and changes the integral to dy/(1-y^4).



Write this out in partial fractions with denominators (1+y^2), (1+y) and (1-y). The numerators are 1/2, 1/4, and 1/4.



The first is a standard one, integrating to arctan(y), while the others are easy as well, giving log(1+y) and -log(1-y) respectively.



Then we have the final answer (1/4)arctan(x^2) + (1/8)log(1+x^2) - (1/8)log(1-x^2)



If you know complex numbers, you can integrate x/(1-x^k) pretty easily for any k, getting the general result



-(1/k)(sum (w_j)^2 log(x-w_j))



where the w_j are the k'th (complex) roots of unity and the sum is over all k of them.



It is a little complicated to take this formula for k = 8 and transform it to what I derived above involving arctan, but I have done it and it works out nicely.

Answer 2

Put x^2=t, so 2x dx = dt. Thus the integral becomes int{dt/(1-t^4)}.Now use partial fractions to get [1/(1-t^2) + 1/(1+t^2)]/2. The formula for int[dt/(1-t^2)] is -ln{(t-1)/(t+1)}/2 and for 1+t^2 is arctan (t). Put t=x^2 and get the answer.

Answer 3

hi bill ok, the two the solutions arrived at are one and the comparable ya. the two might desire to be simplified as ? e^x sin x dx = [e^x (sin x - cos x)] / 2 ?sin(x)e^x dx= -(e^x)cos(x) + [(e^x)(sin(x)) - ?sin(x)e^x dx] After simplification, it fairly is going to become as you wrote 2?sin(x)e^x dx= -(e^x)cos(x) + (e^x)(sin(x)) better simplified into 2?sin(x)e^x dx= (e^x)[(sin(x) - cos x)] So ? e^x sin x dx = [e^x (sin x - cos x)] / 2 surprising. you have written the stairs so of course. yet basically made a slip interior the previous couple of steps. Congrats and each and all the excellent ya.