Integrate x^7/ x+1?

∫{x⁷ / (x + 1)} dx
= ∫{[(x⁷ + x⁶) - (x⁶ + x⁵) + (x⁵ + x⁴) - (x⁴ + x³) + (x³ + x²) - (x² + x) + (x + 1) - 1] / (x + 1)} dx
= ∫{[x⁶(x + 1) - x⁵(x + 1) + x⁴(x + 1) - x³(x + 1) + x²(x + 1) - x(x + 1) + (x + 1) - 1] / (x + 1)} dx
= ∫{x⁶ - x⁵ + x⁴ - x³ + x² - x + 1 - [1/(x + 1)]} dx
= ∫x⁶ dx - ∫x⁵ dx + ∫x⁴ dx - ∫x³ dx + ∫x² dx - ∫x dx + ∫dx - ∫[1/(x + 1)] d(x + 1)
= (x⁷/7) - (x⁶/6) + (x⁵/5) - (x⁴/4) + (x³/3) - (x²/2) + x - ln|x + 1| + C

Meaningless without a differential. OK, I'll pretend you wrote
x^7 dx/(x+1).

Let u = x+1, du = dx, and x^7 = (u-1)^7
= u^7 - 7u^6 + 21u^5 - 35u^4 + 35u^3 - 21u^2 + 7u - 1.
So you'll be integrating
(u^6 - 7u^5 + 21u^4 - 35u^3 + 35u^2 - 21u + 7 - (1/u)) du.
After integration you have
(1/7)u^7 - (7/6)u^5 + (21/5)u^5 - (35/4)u^4 + (35/3)u^3 - (21/2)u^2 + 7u - ln|u| + C.
Now replace "u" with (x+1), and you'll be finished.

Hello,

∫ [x⁷ /(x + 1)] dx =

let:

x + 1 = u

x = u - 1

dx = d(u - 1)

dx = du

yielding, by substitution:

∫ [x⁷ /(x + 1)] dx = ∫ [(u - 1)⁷ /u] du =

(to expand (u - 1)⁷, let's split it into (u - 1)³ (u - 1)² (u - 1)²)

∫ {[(u - 1)³ (u - 1)² (u - 1)²] /u} du =

∫ { {[u³ - (3 ∙ u² ∙ 1) + (3 ∙ u ∙ 1²) - 1³] [u² - (2 ∙ u ∙ 1) + 1²] [u² - (2 ∙ u ∙ 1) +
1²]} /u} du =

∫ {[(u³ - 3u² + 3u - 1) (u² - 2u + 1) (u² - 2u + 1)] /u} du =

∫ {[(u³ - 3u² + 3u - 1) (u⁴ - 2u³ + u² - 2u³ + 4u² - 2u + u² - 2u + 1)] /u} du =

∫ {[(u³ - 3u² + 3u - 1) (u⁴ - 4u³ + 6u² - 4u + 1)] /u} du =

∫ [(u⁷ - 4u⁶ + 6u⁵ - 4u⁴ + u³ - 3u⁶ + 12u⁵ - 18u⁴ + 12u³ - 3u² + 3u⁵ - 12u⁴ + 18u³ -
12u² + 3u - u⁴ + 4u³ - 6u² + 4u - 1) /u] du =

∫ [(u⁷ - 7u⁶ + 21u⁵ - 35u⁴ + 35u³ - 21u² + 7u - 1) /u] du =

(distributing and simplifying)

∫ {(u⁷/u) - [(7u⁶)/u] + [(21u⁵)/u] - [(35u⁴)/u] + [(35u³)/u] - [(21u²)/u] + [(7u)/u] -
(1/u)} du =

∫ [u⁶ - 7u⁵ + 21u⁴ - 35u³ + 35u² - 21u + 7 - (1/u)] du =

(splitting into separate integrals and factoring constants out)

∫ u⁶ du - 7 ∫ u⁵ du + 21 ∫ u⁴ du - 35 ∫ u³ du + 35 ∫ u² du - 21 ∫ u du + 7 ∫ du -
∫ (1/u) du =

[1/(6+1)] u⁶ ⁺ ¹ - 7 [1/(5+1)] u⁵ ⁺ ¹ + 21 [1/(4+1)] u⁴ ⁺ ¹ - 35 [1/(3+1)] u³ ⁺ ¹ +
35 [1/(2+1)] u² ⁺ ¹ - 21 [1/(1+1)] u¹ ⁺ ¹ + 7u - ln | u | + C =

(1/7)u⁷ - 7(1/6)u⁶ + 21(1/5)u⁵ - 35(1/4)u⁴ + 35(1/3)u³ - 21(1/2)u² + 7u -
ln | u | + C =

(1/7)u⁷ - (7/6)u⁶ + (21/5)u⁵ - (35/4)u⁴ + (35/3)u³ - (21/2)u² + 7u - ln | u | + C =

finally let's substitute back (x + 1) for u:

(1/7)(x + 1)⁷ - (7/6)(x + 1)⁶ + (21/5)(x + 1)⁵ - (35/4)(x + 1)⁴ + (35/3)(x + 1)³ -
(21/2)(x + 1)² + 7(x + 1) - ln |x + 1| + C =

(1/7)(x + 1)⁷ - (7/6)(x + 1)⁶ + (21/5)(x + 1)⁵ - (35/4)(x + 1)⁴ + (35/3)(x + 1)³ -
(21/2)(x + 1)² + 7x + 7 - ln |x + 1| + C =

being 7 just a constant we can include it in C, ending with:


(1/7)(x + 1)⁷ - (7/6)(x + 1)⁶ + (21/5)(x + 1)⁵ - (35/4)(x + 1)⁴ + (35/3)(x + 1)³ -
(21/2)(x + 1)² + 7x - ln |x + 1| + C



I hope it's helpful

Don't make things complicated. If you used substitution u = x + 1, you should expand (x + 1)⁷, (x + 1)⁶ ...... (x + 1)² and then simplified the result. Otherwise, the calculation is not complete.