
\begin{enumerate}
\item $I = \dint_1^2 x\sqrt{x}\dx$. Posons : $\begin{aligned}[t]
u(x) = \sqrt{x} & \qquad u'(x)=\dfrac{1}{2\sqrt{x}}\\
v'(x) = x & \qquad v(x) = \dfrac{1}{2}x^2
\end{aligned}$

Alors,
\begin{align*}
I & = [(uv)(x)]_1^2 - \dint_1^2 u'(x)v(x)\dx\\
& = \left[\dfrac{1}{2}x^2\sqrt{x}\right]_1^2 - \dint_1^2\dfrac{1}{4}\dfrac{x^2}{\sqrt{x}}\dx\\
& = \left(\dfrac{1}{2}(2)^2\sqrt2 - \dfrac{1}{2}(1)^2\sqrt1\right)-\dfrac{1}{4}\dint_1^2x\sqrt{x}\dx\\
I & = 2\sqrt2-\dfrac{1}{2}-\dfrac{1}{4}I\\
\dfrac{5}{4}I & = 2\sqrt2-\dfrac{1}{2}\\
I & = \dfrac{4}{5}\left(2\sqrt2-\dfrac{1}{2}\right)\\
\Aboxed{I & = \dfrac{8\sqrt2-2}{5}}
\end{align*}



\item $K = \dint_1^\e x\ln(x)\dx$. Posons:

\begin{multicols}{2}
$u(x) = \ln(x)$

$v'(x) = x$

\columnbreak

$u'(x)=\dfrac{1}{x}$

$v(x) = \dfrac{1}{2}x^2$
\end{multicols}

Alors,
\begin{align*}
K & = [(uv)(x)]_1^\e - \dint_1^\e u'(x)v(x)\dx\\
  & = \left[\dfrac{1}{2}x^2\ln(x)\right]_1^\e - \dfrac{1}{2}\dint_1^\e x \dx\\
  & = \dfrac{1}{2}\e^2 - \dfrac{1}{2}\left[\dfrac{1}{2}x^2\right]_1^\e\\& = \dfrac{1}{2}\e^2 - \dfrac{1}{4}(\e^2-1)\\
\Aboxed{K  & = \dfrac{\e^2+1}{4}}
\end{align*}

\item $L = \dint_1^\e x\big(\ln(x)\big)^2\dx$. Posons:

\begin{multicols}{2}
$u(x) = \big(\ln(x)\big)^2$

$v'(x) = x$

\columnbreak

$u'(x)=2\times\ln(x)\times\dfrac{1}{x}$

$v(x) = \dfrac{1}{2}x^2$
\end{multicols}

Alors,
\begin{align*}
L & = [(uv)(x)]_1^\e - \dint_1^\e u'(x)v(x)\dx\\
& = \left[\dfrac{1}{2}x^2\ln^2(x)\right]_1^\e - \underbrace{\dint x\ln(x) \dx}_{=K}\\
& = \dfrac{1}{2}\e^2 - \dfrac{\e^2+1}{4}\\
\Aboxed{L & = \dfrac{\e^2-1}{4}}
\end{align*}
\end{enumerate}
