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

1. amistre64

factor out the bottom

2. amistre64

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

3. amistre64

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

4. amistre64

pencil and paper might help me out...

5. anonymous

use partial fractions with that factored bottom

6. anonymous

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

7. amistre64

its been awhile since I tried decomposing fractions lol

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. amistre64

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

13. anonymous

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

14. amistre64

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

15. amistre64

might be a (-) that I missed

16. anonymous

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)

17. anonymous

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

18. anonymous

so the whole antiderivative will be $\ln x -\ln \sqrt{x^2+1}$ +C or combining more $\ln x/\sqrt{x^2+1}$ +c