## anonymous 5 years ago complete the following integral or partial substitution with a. ∫(x^5+2x)/(x^6+6x^2+56)^2 dx

1. anonymous

∫(x^5+2x)/(x^6+6x^2+56)^2 dx ∫(x^5+2x) dx - ∫(x^6+6x^2+56)^-2 dx

2. anonymous

You can't do this in products, only in additions. suzi20

3. anonymous

u = x^6 +6x^2 +56 du = 6x^5 +12x dx du = 6(x^5 +2x) dx

4. anonymous

dv = (x^6+6x^2+56)^-2 dx v = -1 (x^6+6x^2+56)^-1 uv - ∫ v du (x^6 +6x^2 +56)(-(x^6+6x^2+56)^-1) - 1/6∫(-(x^6+6x^2+56)^-1) (x^5 +2x) dx hmm... hard

5. anonymous

6. anonymous

u have to do partial again

7. anonymous

Let x^6 + 6x^2 + 56 = t Now, du = 6x^5 + 12x dx = 6(x^5 + 2x) dx The integral becomes, $1/6\int\limits_{}{}dt/t^2$ $-1/6t$ Putting back the value of t = -1/6(x^6+6x^2+56)

8. anonymous

I could ask you to complete

9. anonymous

that's calculus 2, i remember now, wow amogh cool...........

10. anonymous

amogh can I ask you to write down the answer from the first until the end if you're willing I will be very thankful to you

11. anonymous

he did

12. anonymous

Tell me what you didn't understand, I've written it completely!

13. anonymous

oh I'm sorry, I really do not understand about the calculus lesson, but I want to learn

14. anonymous

Let x^6 + 6x^2 + 56 = t Differentiating on both sides, Now, dt = (6x^5 + 12x)dx = 6(x^5 + 2x) dx Multiplying the integral by 6 and diving by 6, $1/6\int\limits_{}^{}6(x^5+2x)/(x^6+6x^2+56)^2 dx$ Now 6(x^5+2x) becomes dt. The integral becomes, $1/6\int\limits_{}^{} dt/t^2$ = −1/6t Putting back the value of t = -1/6(x^6+6x^2+56)

15. anonymous

amogh thanks a lot