antiderivative of 1/((x^3)+x) ....please show steps

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antiderivative of 1/((x^3)+x) ....please show steps

Mathematics
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factor out the bottom
1 ---------- ..... maybe that doesnt help as much :) x(x^2 + 1)
try this... u = x^3 + x du = 3x^2 +1 dx does that help any?

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pencil and paper might help me out...
use partial fractions with that factored bottom
set it to A/x +(Bx+C)/(x^2+1)
its been awhile since I tried decomposing fractions lol
yeah, lol. I will try to explain it best I can
you multiple both sides of the equation by x(x^2+1)
this leaves you with 1= A(x^2+1) and (Bx+C)(x) which when you distribute gives you 1=Ax^2 +A+Bx^2+CX
now group up the like terms on the right hand side 1= (A+B)x^2+Cx+A using the method of undetermined coefficients, now set each coefficient on the right hand side equal the one on the left (filling in zeroes for missing terms) 0x^2+0x+1=(A+B)x^2+Cx+A this means A+B=0 c=0 and A=1 are the three eqs. hence, A and C are done right off the bat and substituting A=1 into the A+B=0 gives 1+B=0 and B=-1
.....ouch......my brain is hurting :) youre doing a great job at this tho
Now going back to the original partial fraction decompostion 1/x(x^2+1)= A/x + (Bx+C)/(x^2+1) gives us 1/x + -x/(x^2+1) now we integrate each of these the first integral is ln(x), but the second we use a u-substitution with u=x^2+1
(1/2) ln(x^2+1)... :)
might be a (-) that I missed
therefore du=2xdx, but up top we only have a -x, so we need to pull out the -1 and multiply by 2 (and then out front of the integral also divide by 2 so it will look like -1/2* int(2x/(x^2+1))= -1/2 int(du/u) integrating this gives -1/2 ln(u) and then substituting back in gives - 1/2 ln (x^2+1)
yeah, the - still out front, but you got it :)
so the whole antiderivative will be \[\ln x -\ln \sqrt{x^2+1}\] +C or combining more \[\ln x/\sqrt{x^2+1}\] +c

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