Evaluate $\int_{2}^{3}\frac{(x-2)^{100}}{(x-1)}dx$ Evaluate $$\int_{2}^{3}\frac{(x-2)^{100}}{(x-1)}dx$$
Is there an easy method to evaluate the integral. I wanted to try by using long division but I think that it will be very complicate. Any help ,thanks   
 A: The long division takes you
$$\frac{(x-2)^{100}}{(x-2)+1}=(x-2)^{99}-(x-2)^{98}+(x-2)^{97}-......+(x-2)-1+\frac{1}{x-1}$$
so
$$\int \frac{(x-2)^{100}}{(x-1)}dx=\frac{(x-2)^{100}}{100}-\frac{(x-2)^{99}}{99}+.......-x+\log(x-1)$$
$$\int_{2}^{3}\frac{(x-2)^{100}}{(x-1)}dx=\frac{1}{100}-\frac{1}{99}+\frac{1}{98}.......-1+\log(2)$$
$$=\log(2)-\frac{47979622564155786918478609039662898122617}{69720375229712477164533808935312303556800}$$
A: We can easily generalize the development for 
$$I_n=\int_2^3 \frac{(x-2)^n}{(x-1)}dx \tag 1$$
We write 
$$I_n=\int_2^3 \frac{(x-2)^n}{(x-1)}dx=\int_1^2 \frac{(x-1)^n}{x}dx=\int_1^2 \frac{x-1}{x}(x-1)^{n-1}dx=\frac{1}{n}-I_{n-1}$$
Note that $I_0=\log 2$ and $I_n=\frac{1}{n}-I_{n-1}$.  So, we simply iterate to reveal that
$$I_1=-\log 2+1$$
$$I_2=\log 2 -1+\frac12$$
$$I_3=-\log 2+1-\frac12+\frac13$$
$$\begin{align}
I_n&=(-1)^n(\log 2-1+1/2-1/3+\cdots +1/n)\\\\
&=(-1)^n\left(\log2 +\sum_{k=1}^{n}\frac{(-1)^k}{k}\right) \tag 2
\end{align}$$
For $n=100$, we have 
$$\begin{align}
I_{100}&=\log 2+\sum_{k=1}^{100}\frac{(-1)^{k}}{k}\\\\
&=\log 2 -\frac{47979622564155786918478609039662898122617}{69720375229712477164533808935312303556800}\\\\
&\approx. 0.004975
\end{align}$$
where the fraction and the decimal approximation were obtained using Wolfram Alpha.

NOTE:
This approach can be used to prove that the value of the alternating harmonic series is indeed $\log 2$.  
Simply observe the limit as $n\to \infty$ of $(1)$ and $(2)$.  For $(1)$, we see that $I_n \to 0$ as $n\to \infty$. 
We also see that  $|I_n| \to |\log 2+\sum_{k=1}^{\infty}\frac{(-1)^n}{n}|$.  
Since the term in absolute value must be zero, then $\sum_{k=1}^{\infty}\frac{(-1)^{n+1}}{n}=\log 2$.
