Question

Question #e7b85

Answer 1

`x^8/8+x^7/7+x^6/6+x^5/5+x^4/4+x^3/3+x^2/2+x+C`

Explanation:

I will begin by splitting the integral up into two:
`int\ (1-x^8)/(1-x)\ dx=int\ 1/(1-x)\ dx-int\ x^8/(1-x)\ dx`

I will call the left one Integral 1 and the right one Integral 2.

Integral 1
To crack this, I will introduce a u-substitution with `u=1-x`, and the derivative being `-1`, so we divide through by `-1` to integrate with respect to `u`:
`int\ 1/(1-x)\ dx=int\ 1/(-1*u)\ du=-int\ 1/u\ du=-ln|u|=-ln|1-x|`

Integral 2
Since the numerator of this integral is of greater degree than the denominator, we use polynomial long division to simplify like so:
`int\ 1/(1-x)-x^7-x^6-x^5-x^4-x^3-x^2-x-1\ dx=`

We already worked out the first bit in Integral 1, and the rest can be computed with the reverse power rule:
`=-ln(1-x)-x^8/8-x^7/7-x^6/6-x^5/5-x^4/4-x^3/3-x^2/2-x`

Completing the original integral
Now that we know the answer to Integral 1 and Integral 2, we can put them together to get the answer for the original integral:
`-ln(1-x)-(-ln(1-x)-x^8/8-x^7/7-x^6/6-x^5/5-x^4/4-x^3/3-x^2/2-x)`

`=-ln(1-x)+ln(1-x)+x^8/8+x^7/7+x^6/6+x^5/5+x^4/4+x^3/3+x^2/2+x`

`=x^8/8+x^7/7+x^6/6+x^5/5+x^4/4+x^3/3+x^2/2+x+C`