Stelling
Als f en g differentieerbaar zijn dan is:
$
\int {f(x)g'\left( x \right)} \,dx = f\left( x \right) \cdot g(x) - \int {g(x) \cdot f'(x)\,dx}
$
Ook wel:
$
\int {f(x)dg(x) = f\left( x \right)g(x) - \int {g(x)df\left( x \right)} }
$
$
\int {x \cdot \ln (x)\,dx = \int {\ln (x) \cdot x\,\,dx = \ln (x) \cdot \frac{1}
{2}} } x^2 - \int {\frac{1}
{2}} x^2 \cdot \frac{1}
{x}\,dx = \frac{1}
{2}x^2 \ln (x) - \frac{1}
{4}x^2
$
$
\int {x \cdot e^x } dx = x \cdot e^x - \int {e^x \cdot 1\,dx = x \cdot e^x } - e^x = (x - 1) \cdot e^x
$
$
\int {x^2 \cdot e^x } dx = x^2 \cdot e^x - \int {e^x \cdot 2x\,dx}
$
Voor dit laatste geldt:
$
\eqalign{
& \int {e^x \cdot 2x\,dx} = 2x \cdot e^x - \int {e^x } \cdot 2\,dx = 2x \cdot e^x - 2e^x \cr}
$
Dus:
$
\int {x^2 \cdot e^x } dx = x^2 \cdot e^x - \left( {2x \cdot e^x - 2e^x } \right) = \left( {x^2 - 2x + 2} \right) \cdot e^x
$
$
\int {\ln (x)\,dx = \int {\ln (x)} } \cdot 1\,dx = \ln (x) \cdot x - \int {x \cdot \frac{1}
{x}} \,dx = x\ln (x) - \int {1\,dx = x\ln (x) - x}
$
$
\eqalign{
& \int {\frac{{\ln \left( x \right)}}
{{x^2 }}} dx = \int {\ln \left( x \right)} \cdot \frac{1}
{{x^2 }}dx = \ln \left( x \right) \cdot \frac{{ - 1}}
{x} - \int {\frac{{ - 1}}
{x} \cdot \frac{1}
{x}} dx = - \frac{{\ln \left( x \right)}}
{x} - \frac{1}
{x} + C \cr
& \int\limits_{x = 1}^e {\frac{{\ln \left( x \right)}}
{{x^2 }}} dx = \left[ { - \frac{{\ln \left( x \right)}}
{x} - \frac{1}
{x}} \right]_1^e = - \frac{{\ln \left( e \right)}}
{e} - \frac{1}
{e} - \left( { - \frac{{\ln \left( 1 \right)}}
{1} - \frac{1}
{1}} \right) = - \frac{2}
{e} + 1 = \frac{{e - 2}}
{e} \cr}
$
$
\int {\sin (x) \cdot e^x } dx = \sin (x) \cdot e^x - \int {e^x \cdot \cos (x)\,dx}
$
Dat schiet niet op?
$
\int {e^x \cdot \cos (x)\,dx} = \cos (x) \cdot e^x - \int {e^x \cdot - \sin (x)\,dx = } \cos (x) \cdot e^x + \int {e^x \cdot \sin (x)} \,dx
$
Lekker...:-)
$
\int {\sin (x) \cdot e^x } dx = \sin (x) \cdot e^x - \cos (x) \cdot e^x - \int {e^x \cdot \sin (x)} \,dx
$
Alhoewel:
$
\eqalign{
& 2 \cdot \int {\sin (x) \cdot e^x } dx = \sin (x) \cdot e^x - \cos (x) \cdot e^x \cr
& \int {\sin (x) \cdot e^x } dx = \frac{1}
{2}e^x \left( {\sin (x) - \cos (x)} \right) \cr}
$