How should I solve this Integral? (x^3 + 3x^2 + 2x + 1)/(x - 1)

[waleed8907]

$$\frac{\left(x^3+3x^2+2x+1\right)}{x-1}$$
My attempt trying to substitute x-1 as u:

 

[nasi112]

Use long division and the integral will be solved straightforward.

 

[Steven G]

You can't solve an integral that has du AND dx.
The integral of a product does not become the sum of two integrals.
If your integral have x's in it and you decide to make a u-substitution you must convert all x's to u's including the dx

 

[khansaheb]

$$\frac{\left(x^3+3x^2+2x+1\right)}{x-1}$$
My attempt trying to substitute x-1 as u:

View attachment 35998

Use long division and the integral will be solved straightforward.

{x^3 + 3x^2 + 2x +1} / {x - 1} = (x^2 + 4x + 6 ) + [7/{x+1}] .......continue

 

[khansaheb]

{x^3 + 3x^2 + 2x +1} / {x - 1} = (x^2 + 4x + 6 ) + [7/{x+1}] .......continue

The above is incorrect. I wrote {x+1} as the denominator in one of the expressions. The correct expression should be:

$\displaystyle \frac {x^3 + 3x^2 + 2x +1} {x - 1}$

= (x^2 + 4x + 6 ) + $\displaystyle \frac{7}{x-1}$ .

I used some redundant parentheses for grouping.

 

[NotJeffM]

Solving this by u-substitution is possible if you do it correctly, but why bother. It OBVIOUSLY will involve a nightmare of error-prone algebra.

$$y = \dfrac{x^3 + 3x^2 + 2x + 1}{x - 1} = \dfrac{(x^3 - x^2) + 4(x^2 - x) + 6(x - 1) + 7}{(x -1)} = \\ x^2 + 4x + 6 + \dfrac{7}{x - 1}.\\ \therefore \ \int y \ dx = \int x^2 \ dx + 4\int x \ dx + \int 6 \ dx + 7 \int \dfrac{dx}{x - 1}$$
It is possible to make an error in algebra in that, but you end up with a sum of very easy integrals.

You can of course do a u-substitution.

$$\text {Set } u = x - 1 \implies dx = du, \ x = u + 1, \text { and }\\ y = \dfrac{(u + 1)^3 + 3(u + 1)^2 + 2(u + 1) + 1}{u} =\\ \dfrac{u^3 + 3u^2 + 3u + 1 + 3u^2 + 6u + 3 + 2u + 2 + 1}{u} =\\ \dfrac{u^3 + 6u^2 + 11u + 7}{u} = u^2 + 6u + 11 + \dfrac{7}{u} =\\ (x - 1)^2 + 6(x - 1) + 11 + \dfrac{7}{x - 1} = \\ x^2 - 2x + 1 + 6x - 6 + 11 + \dfrac{7}{x - 1} =\\ x^2 + 4x + 6 + \dfrac{7}{x - 1}.$$
More work and thus more opportunity for error.

Think before you choose a technique.