A Fancy Derivative

Initially I headed the wrong way since I assumed the last term was ln(u - 1) instead of ln(u) - 1. It would be nice if the question was rewritten to make this less ambiguous.

When finding the derivative, it is important to remember that $n$ is a constant. We are only differentiating in respect to $u$ .

$\frac{d}{du} \left( \frac{u^{n+1}}{(n+1)^2} \cdot \left[ (n+1) \ln u -1 \right] \right)$

= $\frac{d}{du} \frac{u^{n+1}}{(n+1)^2} \cdot \left[ (n+1) \ln u -1 \right] + \frac{u^{n+1}}{(n+1)^2} \cdot \frac{d}{du} \left[ (n+1) \ln u -1 \right]$

= $\frac{(n+1)u^n}{(n+1)^2} \cdot \left[ (n+1) \ln u -1 \right] + \frac{u^{n+1}}{(n+1)^2} \cdot \frac{n+1}{u}$

= $\frac{u^n}{n+1} \cdot \left[ (n+1) \ln u -1 \right] + \frac{u^n}{n+1}$

= $\frac{u^n (n+1) \ln u}{n+1} - \frac{u^n}{n+1} + \frac{u^n}{n+1}$

= $u^n \ln u - \frac{u^n}{n+1} + \frac{u^n}{n+1}$

= $u^n \ln u$

Here the Chain Rule and Product rule of differentiation should be used, n+1 with ln(u) is taken over (u)

solve it by integrating....

See my solution here .

Very simple calculus, I did it in my head. But if you have to work it out just use the General Power Rule, Quotient Rule, and Product rule with some simple pre-calc for the Natural Log.

Diff with u.v rule