How Can I Simplify This Integral Using Long Division?

[tmt1]

I have this integral

$$6\int_{}^{} \frac{u^3 - 1 + 1}{u - 1}\,d$$

And I need to simplify it to

$$6\int_{}^{}u^2 + u + 1 \frac{1}{u - 1}\,du$$

But I don't know how to get to this step.

 

[MarkFL]

The crucial step here is to simplify:

$$\frac{u^3-1}{u-1}$$

What happens if you factor the numerator as the difference of cubes?

 

[soroban]

I have this integral

$$6\int_{}^{} \frac{u^3 - 1 + 1}{u - 1}\,d$$

And I need to simplify it to

$$6\int_{}^{}u^2 + u + 1 + \frac{1}{u - 1}\,du$$

But I don't know how to get to this step.

How about Long Division?

$$\begin{array}{cccccccccc}<br /> & & & & u^2 & +& u &+& 1 \\<br /> & & - & - & - & - & - & - & - \\<br /> u-1 & ) & u^3 \\<br /> & & u^3 & - & u^2 \\<br /> & & - & - & - \\<br /> & & & & u^2 \\<br /> &&&& u^2 &-& u \\<br /> &&&& -&-&- \\<br /> &&&&&& u \\<br /> &&&&&& u &-& 1 \\<br /> &&&&&& - & - & - \\<br /> &&&&&&&& 1 <br /> \end{array}$$$$\text{Therefore: }\;\frac{u^3}{u-1} \;=\;u^2 + u + 1 + \frac{1}{u-1}$$