Integral/Wurzel 3x^2+5x-4/Aufgabe/Lösung

Wir schreiben

{}{\begin{aligned}3x^{2}+5x-4&={\left({\sqrt {3}}x+{\frac {5}{2{\sqrt {3}}}}\right)}^{2}-{\frac {25}{12}}-4\\&={\left({\sqrt {3}}x+{\frac {5}{2{\sqrt {3}}}}\right)}^{2}-{\frac {73}{12}}\\&={\frac {73}{12}}\left(({\frac {{\sqrt {12}}{\sqrt {3}}}{\sqrt {73}}}x+{\frac {5{\sqrt {12}}}{2{\sqrt {3}}{\sqrt {73}}}})^{2}-1\right)\\&={\frac {73}{12}}\left(\left({\frac {6}{\sqrt {73}}}x+{\frac {5}{\sqrt {73}}}\right)^{2}-1\right).\end{aligned}}

Daher ist mit der Substitution

{}u={\frac {6}{\sqrt {73}}}x+{\frac {5}{\sqrt {73}}}\,

bzw.

{}x={\frac {\sqrt {73}}{6}}u-{\frac {5}{6}}\,

{}\int _{}^{}{\sqrt {3x^{2}+5x-4}}\,dx=\int _{}^{}{\sqrt {{\frac {73}{12}}(u^{2}-1)}}\cdot {\frac {\sqrt {73}}{6}}\,du={\frac {73}{12{\sqrt {3}}}}\int _{}^{}{\sqrt {u^{2}-1}}\,du\,.

Eine Stammfunktion hiervon ist

{\frac {73}{12{\sqrt {3}}}}{\frac {1}{2}}(u{\sqrt {u^{2}-1}}-\,\operatorname {arcosh} \,u\,)

und damit ist

{\frac {73}{24{\sqrt {3}}}}\left(({\frac {6}{\sqrt {73}}}x+{\frac {5}{\sqrt {73}}}){\sqrt {({\frac {6}{\sqrt {73}}}x+{\frac {5}{\sqrt {73}}})^{2}-1}}-\,\operatorname {arcosh} \,({\frac {6}{\sqrt {73}}}x+{\frac {5}{\sqrt {73}}})\,\right)

eine Stammfunktion von

{\sqrt {3x^{2}+5x-4}}.