Να υπολογίσετε το ολοκλήρωμα:
$$\mathcal{J}=\int_{0}^{1} \ln^2 \left ( x^2+\sqrt{x^2+1} \right )\, {\rm d}x$$
Λύση
Έχουμε διαδοχικά:
\begin{align*}
\int_{0}^{1}f^2 (x)\, {\rm d}x &= \int_{0}^{1}\ln^2 \left ( x + \sqrt{x^2+1} \right )\, {\rm d}x\\
&= \left [ x \ln^2 \left ( x + \sqrt{x^2+1} \right ) \right ]_0^1 - 2\int_{0}^{1}\frac{x\ln \left ( x+ \sqrt{x^2+1} \right )}{\sqrt{x^2+1}}\, {\rm d}x \\
&=\ln^2 \left ( 1+ \sqrt{2} \right )- \left [2 \sqrt{x^2 +1} \ln \left ( x+ \sqrt{x^2+1} \right ) \right ]_0^1 + 2\int_{0}^{1}\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}\, {\rm d}x \\
&=\ln^2 \left ( 1 + \sqrt{2} \right ) - 2 \sqrt{2}\ln \left ( 1+ \sqrt{2} \right ) + 2\int_{0}^{1}\, {\rm d}x\\
&= \ln^2 \left ( 1+ \sqrt{2} \right ) - 2\sqrt{2} \ln \left ( 1+ \sqrt{2} \right )+ 2
\end{align*}
διότι $\displaystyle{\left ( \ln \left ( x+\sqrt{x^2+1} \right ) \right )' = \frac{1}{\sqrt{x^2+1}}}$.
$$\mathcal{J}=\int_{0}^{1} \ln^2 \left ( x^2+\sqrt{x^2+1} \right )\, {\rm d}x$$
Λύση
Έχουμε διαδοχικά:
\begin{align*}
\int_{0}^{1}f^2 (x)\, {\rm d}x &= \int_{0}^{1}\ln^2 \left ( x + \sqrt{x^2+1} \right )\, {\rm d}x\\
&= \left [ x \ln^2 \left ( x + \sqrt{x^2+1} \right ) \right ]_0^1 - 2\int_{0}^{1}\frac{x\ln \left ( x+ \sqrt{x^2+1} \right )}{\sqrt{x^2+1}}\, {\rm d}x \\
&=\ln^2 \left ( 1+ \sqrt{2} \right )- \left [2 \sqrt{x^2 +1} \ln \left ( x+ \sqrt{x^2+1} \right ) \right ]_0^1 + 2\int_{0}^{1}\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}\, {\rm d}x \\
&=\ln^2 \left ( 1 + \sqrt{2} \right ) - 2 \sqrt{2}\ln \left ( 1+ \sqrt{2} \right ) + 2\int_{0}^{1}\, {\rm d}x\\
&= \ln^2 \left ( 1+ \sqrt{2} \right ) - 2\sqrt{2} \ln \left ( 1+ \sqrt{2} \right )+ 2
\end{align*}
διότι $\displaystyle{\left ( \ln \left ( x+\sqrt{x^2+1} \right ) \right )' = \frac{1}{\sqrt{x^2+1}}}$.
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