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