Problem source

Evaluate the following integrals, in the different cases that arise according to the value of the positive constant $a\,$:
\begin{questionparts}
\item \[ \displaystyle
\int_0^1  \frac 1 {x^2 + (a+2)x +2a} \; \d x \]
\item \[\displaystyle \int _{1}^2\frac 1 {u^2 +au +a-1} \; \d u\]
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$

\begin{align*}
&& I &= \int_0^1  \frac 1 {x^2 + (a+2)x +2a} \; \d x \\
&&&= \int_0^1 \frac{1}{(x+a)(x+2)} \d x\\
\end{align*}

Case 1: $a = 2$

\begin{align*}
&& I &= \int_0^1 \frac{1}{(x+2)^2} \d x \\
&&&= \left [ -(x+2)^{-1}\right]_0^1 = \frac12 - \frac13 = \frac16
\end{align*}

Case 2: $a \neq 2, a \not \in [0,1]$

\begin{align*}
&& I &=\frac{1}{a-2} \int_0^1 \left ( \frac{1}{x+2} - \frac{1}{x+a} \right) \d x \\
&&&= \frac{1}{a-2} \left [ \ln |x+2| - \ln |x + a|\right]_0^1 \\
&&&= \frac{1}{a-2} \left ( \ln \frac{3}{|1+a|} - \ln \frac{2}{|a|} \right) \\
&&&= \frac{1}{a-2} \ln \frac{3|a|}{2|a+1|}
\end{align*}

Case 3: $a \in [0, 1]$, $I$ does not converge

\item $\,$ \begin{align*}
&& J &= \int _{1}^2\frac 1 {u^2 +au +a-1} \; \d u \\
&&&= \int_1^2 \frac{1}{(u+a-1)(u+1)} \d u \\
x = u-1:&&&= \int_0^1 \frac{1}{(x+a)(x+2)} \d x
\end{align*}
So it's the same as the previous integral
\end{questionparts}