Problem source

\begin{questionparts}
\item Evaluate the integral
\[
\int_0^1  \l x + 1 \r ^{k-1} \; \mathrm{d}x
\]
in the cases  $k\ne0$ and $k = 0\,$.
Deduce that  $\displaystyle \frac{2^k - 1}{k} \approx \ln 2$ when $k \approx 0\,$.

\item Evaluate the integral
\[
\int_0^1  x  \l x + 1 \r ^m \; \mathrm{d}x \;
\]
in the different cases that arise according to the value of $m$.
\end{questionparts}

Solution source

\begin{questionparts}
\item Case $k \neq 0$:
\begin{align*}
&& \int_0^1 (x+1)^{k-1} \d x &= \left [\frac{1}{k}(x+1)^k \right]_0^1 \\
&&&= \frac{2^k-1}{k} \\
\end{align*}
Case $k = 0$:
\begin{align*}
&& \int_0^1 (x+1)^{k-1} \d x &= \int_0^1 (x+1)^{-1} \d x \\
&&&= \left [\ln(x+1) \right]_0^1 \\
&&&= \ln 2
\end{align*}

Therefore for $k \approx 0$, we must have both integrals being close to each other, since the function is nice on this interval, ie $\frac{2^k-1}{k} \approx \ln 2$

\item Case $m = 0$. $I = \frac12$

Case $m \neq 0, -1, -2$

\begin{align*}
u = x+1, \d u = \d x && \int_0^1 x(x+1)^m \d x &= \int_{u=1}^{u=2} (u-1)u^m \d u \\
&&&=\left[ \frac{u^{m+2}}{m+2} - \frac{u^{m+1}}{m+1} \right]_1^2  \\
&&&= 2^{m+1}\left (  \frac{2}{m+2} - \frac1{m+1} \right) - \frac{1}{m+2} + \frac{1}{m+1} \\
&&&= 2^{m+1} \frac{m}{(m+1)(m+2)} + \frac{1}{(m+1)(m+2)} \\
&&&= \frac{m2^{m+1}+1}{(m+1)(m+2)} \\
\end{align*}

Case $m = -1$.

\begin{align*}
&& \int_0^1 \frac{x}{x+1} \d x &= \int_0^1 1 - \frac{1}{x+1} \d x \\
&&&= 1 - \ln2 \\
\end{align*}

Case $m = -2$:

\begin{align*}
&& \int_0^1 \frac{x}{(x+1)^2} \d x &= \int_0^1\frac{x+1-1}{(x+1)^2} \d x \\
&&&= \left [ \ln (x+1) +(1+x)^{-1} \right]_0^1 \\
&&&= \ln 2 + \frac12 -  1 \\
&&&= \ln 2 - \frac12
\end{align*}
\end{questionparts}