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anonymous
 5 years ago
antiderivative of 1/((x^3)+x) ....please show steps
anonymous
 5 years ago
antiderivative of 1/((x^3)+x) ....please show steps

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0factor out the bottom

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01  ..... maybe that doesnt help as much :) x(x^2 + 1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0try this... u = x^3 + x du = 3x^2 +1 dx does that help any?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0pencil and paper might help me out...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use partial fractions with that factored bottom

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0set it to A/x +(Bx+C)/(x^2+1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its been awhile since I tried decomposing fractions lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, lol. I will try to explain it best I can

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you multiple both sides of the equation by x(x^2+1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0.....ouch......my brain is hurting :) youre doing a great job at this tho

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now 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 usubstitution with u=x^2+1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(1/2) ln(x^2+1)... :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0might be a () that I missed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0therefore 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, the  still out front, but you got it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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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