\eqalign{\int {\frac{1}
{{V\left( u \right)}}du = } \frac{2}
{{\sqrt { - D} }}\arctan \left( {\frac{{V'\left( u \right)}}
{{\sqrt { - D} }}} \right) + C
}
Waarbij
V(u) = au^2 + bu + c
en
a > 0
.
Voorbeelden
\eqalign{
& \int {\frac{1}
{{x^2 + x + 1}}dx = \frac{2}
{{\sqrt 3 }}\arctan \left( {\frac{{2x + 1}}
{{\sqrt 3 }}} \right) + C} \cr
& \int {\frac{1}
{{x^2 + 2x + 5}}dx = \frac{2}
{4}} \arctan \left( {\frac{{2x + 2}}
{4}} \right) + C = \frac{1}
{2}\arctan \left( {\frac{{x + 1}}
{2}} \right) + C \cr}
Je kunt dan
V(u)
herschrijven:
\eqalign{
& V(u) = au^2 + bu + c \cr
& V(u) = au^2 + bu + \frac{{b^2 }}
{{4a}} + c - \frac{{b^2 }}
{{4a}} \cr
& V(u) = \left( {au^2 + bu + \frac{{b^2 }}
{{4a}}} \right) + \left( {c - \frac{{b^2 }}
{{4a}}} \right) \cr
& V(u) = \left( {\sqrt a \cdot u + \frac{b}
{{2\sqrt a }}} \right)^2 + \left( {c - \frac{{b^2 }}
{{4a}}} \right) \cr
& V(u) = \left( {\frac{{\sqrt a \cdot u \cdot 2\sqrt a }}
{{2\sqrt a }} + \frac{b}
{{2\sqrt a }}} \right)^2 + \left( {\frac{{c \cdot 4a}}
{{4a}} - \frac{{b^2 }}
{{4a}}} \right) \cr
& V(u) = \left( {\frac{{2au}}
{{2\sqrt a }} + \frac{b}
{{2\sqrt a }}} \right)^2 + \left( {\frac{{4ac}}
{{4a}} - \frac{{b^2 }}
{{4a}}} \right) \cr
& V(u) = \left( {\frac{{2au + b^2 }}
{{2\sqrt a }}} \right)^2 + \frac{{4ac - b^2 }}
{{4a}} \cr}
Je krijgt dan:
\eqalign{\int {\frac{1}
{{V(u)}}du = \int {\frac{1}
{{\left( {\frac{{2au + b^2 }}
{{2\sqrt a }}} \right)^2 + \frac{{4ac - b^2 }}
{{4a}}}}} }}
Omdat
D = b^2 - 4ac
kun je schrijven:
\eqalign{
& \int {\frac{1}
{{V(u)}}du = \int {\frac{1}
{{\left( {\frac{{2au + b^2 }}
{{2\sqrt a }}} \right)^2 + \frac{{ - D}}
{{4a}}}}} } \cr
& \int {\frac{1}
{{V(u)}}du = \int {\frac{1}
{{\left( {\frac{{2au + b^2 }}
{{2\sqrt a }}} \right)^2 + \frac{{ - D}}
{{4a}}}} \cdot \frac{{\frac{{4a}}
{{ - D}}}}
{{\frac{{4a}}
{{ - D}}}}du} } \cr
& \int {\frac{1}
{{V(u)}}du = \frac{{4a}}
{{ - D}}\int {\frac{1}
{{\frac{{4a}}
{{ - D}}\left( {\frac{{2au + b^2 }}
{{2\sqrt a }}} \right)^2 + 1}}\,} } du \cr
& \int {\frac{1}
{{V(u)}}du = \frac{{4a}}
{{ - D}}\int {\frac{1}
{{\left( {\frac{{2\sqrt a }}
{{\sqrt { - D} }}\frac{{2au + b^2 }}
{{2\sqrt a }}} \right)^2 + 1}}\,} } du \cr
& \int {\frac{1}
{{V(u)}}du = \frac{{4a}}
{{ - D}}\int {\frac{1}
{{\left( {\frac{{2au + b^2 }}
{{\sqrt { - D} }}} \right)^2 + 1}}\,} } du \cr}
Je herkent nu wel de primitieve:
\eqalign{
& \int {\frac{1}
{{V(u)}}du = \frac{{\sqrt { - D} }}
{{2a}}\frac{{4a}}
{{ - D}}\arctan \left( {\frac{{2au + b^2 }}
{{\sqrt { - D} }}} \right) + C} \cr
& \int {\frac{1}
{{V(u)}}du = \frac{2}
{{\sqrt { - D} }}\arctan \left( {\frac{{2au + b^2 }}
{{\sqrt { - D} }}} \right) + C} \cr}
Er geldt
V'(u) = 2au + b
dus:
\eqalign{\int {\frac{1}
{{V\left( u \right)}}du = } \frac{2}
{{\sqrt { - D} }}\arctan \left( {\frac{{V'\left( u \right)}}
{{\sqrt { - D} }}} \right) + C
}
Waarbij
V(u) = au^2 + bu + c
en
a > 0
.
Als de discriminant niet strikt negatief is dan kan een oplossing gevonden worden door het opsplitsen in partieel breuken.
Voorbeelden
\eqalign{\int {\frac{1}
{{x^2 + 4x + 4}}dx = } \int {\frac{1}
{{\left( {x + 2} \right)^2 }}dx = } - \frac{1}
{{x + 2}} + C}
\eqalign{\int {\frac{1}
{{x^2 - 5x + 6}}dx = } \int {\frac{1}
{{(x - 2)(x - 3)}}dx = } \ln (x - 3) - \ln (x - 2)}
bron: sigma
