Delvis integration
Til integraler af produkter: \(\int u\cdot v'\,dx = u\cdot v - \int v\cdot u'\,dx\)
Delvis integration bruges, når integranden er et produkt af to funktioner — fx \(\int x\cdot e^x\,dx\). Ideen er at flytte differentieringen fra ét led til det andet, så resten bliver simplere.
Valg af u (LIATE-reglen): Vælg u som det led der bliver simplest ved differentiering: Logaritme > Invers trig > Algebraisk (polynomium) > Trig > Eksponentiel. Log slår altid polynomier; polynomier slår altid e-funktioner.
Husk: v' integreres for at finde v — det er her fejl oftest sker.
u = differentieres
v′ = integreres
Formler
Delvis integration
\(\int \textcolor{#3b82f6}{u}\cdot \textcolor{#10b981}{v'}\,dx = \textcolor{#3b82f6}{u}\cdot \textcolor{#10b981}{v} - \int \textcolor{#10b981}{v}\cdot \textcolor{#3b82f6}{u'}\,dx\)
Vigtige resultater
\(\int x\cdot e^{ax}\,dx = \left(\dfrac{x}{a} - \dfrac{1}{a^2}\right)e^{ax} + c\)
\(\int x\cdot\sin(ax)\,dx = -\dfrac{x}{a}\cos(ax) + \dfrac{1}{a^2}\sin(ax) + c\)
\(\int x\cdot\ln(x)\,dx = \dfrac{x^2}{2}\ln(x) - \dfrac{x^2}{4} + c\)
\(\int x\cdot\sin(ax)\,dx = -\dfrac{x}{a}\cos(ax) + \dfrac{1}{a^2}\sin(ax) + c\)
\(\int x\cdot\ln(x)\,dx = \dfrac{x^2}{2}\ln(x) - \dfrac{x^2}{4} + c\)
Eksempler
Eksempel
\(\int \textcolor{#3b82f6}{x}\cdot \textcolor{#10b981}{e^{2x}}\,dx\)
▼
1
Vælg u og v′
\(\textcolor{#3b82f6}{u = x}\) (differentieres nemt) \(\textcolor{#10b981}{v'=e^{2x}}\)
\(\textcolor{#3b82f6}{u = x}\) (differentieres nemt) \(\textcolor{#10b981}{v'=e^{2x}}\)
2
Find u′ og v
\(\textcolor{#3b82f6}{u'=1}\) \(\textcolor{#10b981}{v=\tfrac{1}{2}e^{2x}}\)
\(\textcolor{#3b82f6}{u'=1}\) \(\textcolor{#10b981}{v=\tfrac{1}{2}e^{2x}}\)
3
Indsæt i formlen
\(= \textcolor{#3b82f6}{x}\cdot\textcolor{#10b981}{\tfrac{1}{2}e^{2x}} - \int\textcolor{#10b981}{\tfrac{1}{2}e^{2x}}\cdot \textcolor{#3b82f6}{1}\,dx = \dfrac{x}{2}e^{2x} - \dfrac{1}{4}e^{2x} + c\)
\(= \textcolor{#3b82f6}{x}\cdot\textcolor{#10b981}{\tfrac{1}{2}e^{2x}} - \int\textcolor{#10b981}{\tfrac{1}{2}e^{2x}}\cdot \textcolor{#3b82f6}{1}\,dx = \dfrac{x}{2}e^{2x} - \dfrac{1}{4}e^{2x} + c\)
Eksempel 2
\(\int \textcolor{#3b82f6}{x}\cdot \textcolor{#10b981}{\ln(x)}\,dx\) — LIATE: L > A
▼
1
Vælg u = ln(x) — logaritme er øverst i LIATE
\(\textcolor{#3b82f6}{u=\ln x},\ u'=\tfrac{1}{x}\) \(\textcolor{#10b981}{v'=x},\ v=\tfrac{x^2}{2}\)
\(\textcolor{#3b82f6}{u=\ln x},\ u'=\tfrac{1}{x}\) \(\textcolor{#10b981}{v'=x},\ v=\tfrac{x^2}{2}\)
2
Indsæt i formlen
\(= \ln x \cdot \tfrac{x^2}{2} - \int\tfrac{x^2}{2}\cdot\tfrac{1}{x}\,dx = \tfrac{x^2}{2}\ln x - \int\tfrac{x}{2}\,dx = \textcolor{#ef4444}{\tfrac{x^2}{2}\ln x - \tfrac{x^2}{4} + c}\)
\(= \ln x \cdot \tfrac{x^2}{2} - \int\tfrac{x^2}{2}\cdot\tfrac{1}{x}\,dx = \tfrac{x^2}{2}\ln x - \int\tfrac{x}{2}\,dx = \textcolor{#ef4444}{\tfrac{x^2}{2}\ln x - \tfrac{x^2}{4} + c}\)
Genkend opgavetypen
🔍 Sådan ser den ud til eksamen
- Integralet er et produkt af to funktioner: \(\int u\cdot v'\,dx\)
- Formelsamlingen: \(\int u\cdot v'\,dx = u\cdot v - \int v\cdot u'\,dx\)
- Typisk: \(\int x\cdot e^{ax}\,dx\), \(\int x\cdot\sin(ax)\,dx\), \(\int x\cdot\ln(x)\,dx\)
- Del 2: kan kræve to omgange delvis integration (\(\int x^2 e^x\,dx\))
⚠️ Klassiske eksamensfejl
- Vælg u som det led der bliver simplere ved differentiering: polynom > ln > sin/cos > e^x
- Glem ikke at integrere v' for at finde v — fejlen sker ofte her
- \(\int x\ln(x)\,dx\): vælg \(u=\ln(x)\), \(v'=x\) — ikke omvendt
- Bestemt delvis integration: husk at indsætte grænser i HELE udtrykket \([uv]_a^b\)
💡 Vælg u som det led der bliver simplere ved differentiering. Typisk: polynom > ln > sin/cos > e^x. Vælg v′ som det resterende led.