Question

Basic Intecal Problem

Answer 1

\[\int\limits_{?}^{?}\frac{ x^{3}dx }{ x-1}\]

Answer 2

there's a particularly nice substitution for that ugly denominator. there are multiple ways to do this, but i would substitute a squared trig function for x-1

Answer 3

Let u = x-1 \[ u= x-1\implies x = u+1\]Now just substitute

Answer 4

\[\int \frac{(u+1)^3}{u}du\]

Answer 5

I think its better if we divide x^3 by x-1

Answer 6

then use power rule, my problem is that I'm not sure how to divide them

Answer 7

When you expand this, you get \[u^3 +3u^2+3u+1\] So we can rewrite the integral as \[\int\frac{u^3 +3u^2+3u+1}{u}du=\int (u^2+3u+4)du\]

Answer 8

After you're done integrating, just resubstitute what u is back into your function of you to get it in terms of x, and that will be your answer.

Answer 9

@Jhannybean we have the same answer. thank you. But I would really appreciate if you teach me how to divide a term by another term

Answer 10

Alright

Answer 11

Ask yourself, what can I multiply to x to turn it into x^3?

Answer 12

x^2

Answer 13

exactly.

Answer 14

So now you multiply x * x^2 and -1* x^2 and write it on the line uderneath.

Answer 15

Now i remember thank you

Answer 16

Yes, so since you have a remainder of -1, you write it like \[x^2 -x +1 -\frac{1}{x-1}\]

Answer 17

@Jhannybean yeap i just wrote that down xD thank you again

Answer 18

No problemo :)

Answer 19

this is a pretty neat formula
\[\rm x^n-1 = (x-1)(1+x+x^2+x^3+\cdots +x^{n-1})\]