5. Primitiveren van wortelvormen

Voorbeeld 1

p1486img9.gif \eqalign{   & \int {\frac{{\sqrt {4 - x^2 } }} {x}} \,dx =   \cr   & Neem:  \cr   & x = 2\sin u \to \sin u = \frac{x} {2}  \cr   & \sqrt {4 - x^2 }  = 2\cos u = \frac{{dx}} {{du}}  \cr   & \int {\frac{{\sqrt {4 - x^2 } }} {x}} \,dx =   \cr   & \int {\frac{{2\cos u}} {{2\sin u}}d(2\sin u) = }   \cr   & \int {\frac{{2\cos u}} {{2\sin u}} \cdot 2\cos u\,\,du = }   \cr   & 2\int {\frac{{\cos ^2 u}} {{\sin u}}du = }   \cr   & 2\int {\frac{{1 - \sin ^2 u}} {{\sin u}}} \,du =   \cr   & 2\int {\left( {\frac{1} {{\sin u}} - \sin u} \right)} \,du =   \cr   & 2\ln \frac{{1 - \cos u}} {{\left| {\sin u} \right|}} + 2\cos u + K =   \cr   & 2\ln \frac{{1 - \frac{1} {2}\sqrt {4 - x^2 } }} {{\left| {\frac{1} {2}x} \right|}} + \sqrt {4 - x^2 }  + K  \cr   & 2\ln \frac{{2 - \sqrt {4 - x^2 } }} {{\left| x \right|}} + \sqrt {4 - x^2 }  + K \cr}


Voorbeeld 2

\eqalign{   & \int {\frac{{\sqrt {2x + 1} }} {{x + 4}}} \,dx?  \cr   & Neem\,\,u = \sqrt {2x + 1}  \to x = \frac{1} {2}u^2  - \frac{1} {2}  \cr   & \int {\frac{{\sqrt {2x + 1} }} {{x + 4}}} dx = \int {\frac{u} {{\frac{1} {2}u^2  - \frac{1} {2} + 4}}} d\left( {\frac{1} {2}u^2  - \frac{1} {2}} \right) = \int {\frac{u} {{\frac{1} {2}u^2  + 3\frac{1} {2}}}} u\,du =   \cr   & \int {\frac{{2u^2 }} {{u^2  + 7}}} \,du = \int {\left( {2 - \frac{{14}} {{u^2  + 7}}} \right)du = } 2u - 2\sqrt 7 \arctan \left( {\frac{{\sqrt 7  \cdot u}} {7}} \right) =   \cr   & 2\sqrt {2x + 1}  - 2\sqrt 7 \arctan \left( {\frac{{\sqrt 7 \sqrt {2x + 1} }} {7}} \right) \cr}

Voorbeeld 3

\eqalign{   & \int {\frac{{\sqrt {2x + 1} }} {{x - 4}}} \,dx?  \cr   & Neem\,\,u = \sqrt {2x + 1}  \to x = \frac{1} {2}u^2  - \frac{1} {2}  \cr   & \int {\frac{{\sqrt {2x + 1} }} {{x - 4}}} \,dx = \int {\frac{u} {{\frac{1} {2}u^2  - \frac{1} {2} - 4}}} d\left( {\frac{1} {2}u^2  - \frac{1} {2}} \right) = \int {\frac{{2u^2 }} {{u^2  - 9}}du = }   \cr   & \int {\left( {\frac{3} {{u - 3}} - \frac{3} {{u + 3}} + 2} \right)du = 3\ln (u - 3) - 3\ln (u + 3) + 2u = }   \cr   & 3\ln \frac{{u - 3}} {{u + 3}} + 2u = 3\ln \frac{{\sqrt {2x + 1}  - 3}} {{\sqrt {2x + 1}  + 3}} + 2\sqrt {2x + 1}  \cr}


Voorbeeld 4

\eqalign{   & \int {\sqrt {x^2  - 8x + 25} } \,dx = \int {\sqrt {(x - 4)^2  + 9} } \,dx =   \cr   & {\text{Neem}}\,\,{\text{t = x - 4}}  \cr   & \int {\sqrt {t^2  + 9} } \,dt = \int {\sqrt {t^2  + 3^2 } } \,dt =   \cr   & {\text{standaardintegraal}}\,\,{\text{1}}{\text{.25}}  \cr   & \frac{t} {2}\sqrt {3^2  + t^2 }  + \frac{{3^2 }} {2}\ln \left( {t + \sqrt {3^2  + t^2 } } \right) + C =   \cr   & \frac{{\left( {x - 4} \right)}} {2}\sqrt {3^2  + \left( {x - 4} \right)^2 }  + \frac{9} {2}\ln \left( {x - 4 + \sqrt {3^2  + \left( {x - 4} \right)^2 } } \right) + C =   \cr   & \frac{{\left( {x - 4} \right)}} {2}\sqrt {x^2  - 8x + 25}  + \frac{9} {2}\ln \left( {x - 4 + \sqrt {x^2  - 8x + 25} } \right) + C \cr}

F.A.Q.

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