Problem source

Show that (for $t>0$)
\begin{questionparts}
\item
\[
\int_0^1 \frac1{(1+tx)^2} \d x = \frac1{(1+t)}
\]
\item
\[
\int_0^1 \frac{-2x}{(1+tx)^3} \d x = -\frac1{(1+t)^2}
\]
\end{questionparts}
Noting that the right hand side of (ii) is the derivative of the right hand side of 
(i), 
conjecture the value of
\[
 \int_0^1 \frac{6x^2}{(1+x)^{4}} \d x \;.
\]
(You need not verify your conjecture.)

Solution source

\begin{questionparts}
\item For the first one, consider
\begin{align*}
&& \int_0^1 \frac{1}{(1+tx)^2} \d x &= \left [ -\frac{1}{t}(1+tx)^{-1} \right]_0^1 \\
&&&= \frac{1}{t} - \frac{1}{t(1+t)} \\
&&&= \frac{t+1-1}{t(t+1)} = \frac{1}{t+1}
\end{align*}

\item Consider

\begin{align*}
&& \int_0^1 \frac{-2x}{(1+tx)^3} \d x &= \int_0^1 \frac{\frac{2}{t}(1+tx) -\frac{2}{t}}{(1+tx)^3} \d x \\
&&&= -\frac{2}{t} \int_0^1 \left (\frac{1}{(1+tx)^2}- \frac{1}{(1+tx)^3}  \right) \d x \\
&&&= -\frac{2}{t} \frac{1}{t+1} + \frac{2}{t} \left [ \frac{1}{-2t}(1+tx)^{-2}\right]_0^1 \\
&&&= -\frac{2}{t(t+1)} + \frac2t\left (\frac{1}{2t} - \frac{1}{2t(1+t)^2} \right) \\
&&&= -\frac{2}{t} \left ( \frac{1}{t+1} + \frac{1}{2t(1+t)^2} - \frac{1}{2t}\right) \\
&&&= -\frac{2}{t} \frac{2t(1+t)+1-(1+t)^2}{2t(1+t)^2} \\
&&&= -\frac{2}{t} \frac{2t^2+2t+1-1-2t-t^2}{2t(1+t)^2} \\
&&&= -\frac{1}{(1+t)^2}
\end{align*}
\end{questionparts}

I would expect it to be $\frac{2}{(1+t)^3}$. This is actually an application of differentiating under the integral sign and is completely valid where functions are well behaved.