Stelling
Als F een primitieve is van f op (a,b) en g:(c,d)$\to$(a,b) differentieerbaar is op (c,d) met afgeleide g', dan is:
$\int{}$f(g(x))·g'(x)dx=F(g(x))+K op (c,d)
Ook wel: $\int{}$f(g(x))dg(x)=F(g(x))+K op (c,d)
Toepassing
We zoeken $\int{}$h(x)dx
We noteren:
$\int{}$h(x)dx=$\int{}$f(g(x))g'(x)dx=$\int{}$f(g(x))dg(x)=$\int{}$f(t)dt=F(t)+K=F(g(x))+K (t=g(x))
We hebben dus g(x)=t gesteld, vandaar de naam substitutiemethode.
$
\eqalign{
& \int {x(5 + x^2 )^3 dx = } \cr
& Schijf\,\,h(x)\,\,als\,\,f(g(x) \cdot g'(x)) \cr
& \int {\frac{{\text{1}}}
{{\text{2}}}{\text{(5 + x}}^{\text{2}} {\text{)}}^{\text{3}} \cdot 2x} \,dx = \int {\frac{{\text{1}}}
{{\text{2}}}{\text{(5 + x}}^{\text{2}} {\text{)}}^{\text{3}} \,d\left( {{\text{5 + x}}^{\text{2}} } \right)} = \cr
& Neem\,\,t = 5 + x^2 \cr
& \int {\frac{1}
{2}t^3 } dt = \frac{1}
{8}t^4 + C = \cr
& Neem\,\,t = 5 + x^2 \cr
& \frac{1}
{8}\left( {{\text{5 + x}}^{\text{2}} } \right)^4 + C \cr}
$
$
\eqalign{
& \int {\frac{t}
{{1 + t^2 }}} \,dt = \int {\frac{1}
{2}} \cdot \frac{1}
{{1 + t^2 }} \cdot 2t\,dt = \int {\frac{1}
{2} \cdot \frac{1}
{{1 + t^2 }}} \,d(1 + t^2 ) \cr
& Neem\,\,u = 1 + t^2 : \cr
& \int {\frac{1}
{2}} \cdot \frac{1}
{u}\,du = \frac{1}
{2}\ln (u) + C = \frac{1}
{2}\ln (1 + t^2 ) + C \cr}
$
$
\eqalign{\int {\frac{{\ln (x)}}
{x}} \,dx = \int {\ln (x) \cdot \frac{1}
{x}} \,dx = \int {\ln (x)\,d(\ln (x)) = \int {t\,dt = \frac{1}
{2}} } t^2 + C = \frac{1}
{2}(\ln (x))^2 + C}
$