Question

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

Answer 1

factor out the bottom

Answer 2

1
---------- ..... maybe that doesnt help as much :)
x(x^2 + 1)

Answer 3

try this...

u = x^3 + x
du = 3x^2 +1 dx does that help any?

Answer 4

pencil and paper might help me out...

Answer 5

use partial fractions with that factored bottom

Answer 6

set it to A/x +(Bx+C)/(x^2+1)

Answer 7

its been awhile since I tried decomposing fractions lol

Answer 8

yeah, lol. I will try to explain it best I can

Answer 9

you multiple both sides of the equation by x(x^2+1)

Answer 10

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

Answer 11

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

Answer 12

.....ouch......my brain is hurting :) youre doing a great job at this tho

Answer 13

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

Answer 14

(1/2) ln(x^2+1)... :)

Answer 15

might be a (-) that I missed

Answer 16

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)

Answer 17

yeah, the - still out front, but you got it :)

Answer 18

so the whole antiderivative will be \[\ln x -\ln \sqrt{x^2+1}\] +C

or combining more \[\ln x/\sqrt{x^2+1}\] +c