Super long integration by tabular method???

Question

integratetion of x^13 (sqrtx^7-1)We have tried everything. What do you think. I have set it up several different ways. If it is tablular method, I think it is so long and tedious, that to many chances for mistakes. What do you think?

Answer 1

You've tried almost everything.

Integrate ∫[x^13 √(x^7 - 1)] dx.

Let
w = x^7
dw = 7x^6 dx
(1/7)w dw = x^13 dx

∫[x^13 √(x^7 - 1)] dx

= (1/7)∫[w√(w - 1)] dw

Let
u = w - 1
du = dw
w = u + 1

(1/7)∫[w√(w - 1)] dw

= (1/7)∫[(u + 1)√u] du

= (1/7)∫[u^(3/2) + √u] du

= (1/7) [(2/5)u^(5/2) + (2/3)u^(3/2)] + C

= (2/105) u^(3/2) [3u + 5] + C

= (2/105) (w - 1)^(3/2) [3(w - 1) + 5] + C

= (2/105) (w - 1)^(3/2) [3w + 2] + C

= (2/105) (x^7 - 1)^(3/2) [3x^7 + 2] + C

Answer 2

I think you should multiply x^13 (sqrtx^7-1) out (provided only x^7 is under the square root). Then you have

x^13 · x^(7/2) - x^13 =
x^(33/2) - x^13

which is quite easy to integrate - the integral of this is
2/35 · x^(35/2) - 1/14 · x^14

____

If, however, you have sqrt(x^7-1), then I am afraid one needs to integrate by parts 13 times...

Answer 3

Which do you mean?

∫x^13 (√x^7 - 1) dx or ∫x^13 √(x^7 - 1) dx

If the former, note that √x^7 = x^(7/2), so you have:

∫x^13x^(7/2) - x^13 dx
∫x^(33/2) - x^13 dx
2x^(35/2)/35 - x^14/14 + C

If the latter, make the substitution u=x^7-1, du=7x^6 dx, dx=du/(7x^6). Then you have:

1/7 ∫x^7 √u du

But x^7 = u+1, so:

1/7 ∫(u+1) √u du
1/7 ∫u^(3/2) + u^(1/2) du
1/7 (2u^(5/2)/5 + 2u^(3/2)/3) + C
2u^(5/2)/35 + 2u^(3/2)/21 + C
2(x^7-1)^(5/2)/35 + 2(x^7-1)^(3/2)/21 + C

Answer 4

... What? Are any of those lines supposed to have any relationship with any of those other lines? Because if so, you should probably state that.