回復 5# 阿光 的帖子
第8題
\(\begin{align}
& y={{x}^{2}}-4x+5 \\
& y'=2x-4 \\
& \\
& u=2x-4,du=2dx \\
& \int{\sqrt{{{\left( 2x-4 \right)}^{2}}+1}dx} \\
& =\frac{1}{2}\int{\sqrt{{{u}^{2}}+1}du} \\
& =\frac{1}{2}\left( \frac{u}{2}\sqrt{{{u}^{2}}+1}+\frac{1}{2}\ln \left| u+\sqrt{{{u}^{2}}+1} \right| \right)+C \\
& =\frac{\left( x-2 \right)\sqrt{{{\left( 2x-4 \right)}^{2}}+1}}{2}+\frac{\ln \left| 2x-4+\sqrt{{{\left( 2x-4 \right)}^{2}}+1} \right|}{4}+C \\
& \\
& \int_{1}^{2}{\sqrt{{{\left( 2x-4 \right)}^{2}}+1}dx=0-\left( -\frac{\sqrt{5}}{2}+\frac{\ln \left( \sqrt{5}-2 \right)}{4} \right)=}\frac{\sqrt{5}}{2}-\frac{\ln \left( \sqrt{5}-2 \right)}{4} \\
\end{align}\)
