Integraal is te vinden op The Daily Integral 1-3-2026.

\[ \int_{0}^{1}\frac{x^{2}\ln\left(x^{10}+x^{9}+x+1\right)}{1+x^{3}}dx \]

Factoriseren en lineariteit toepassen

\[ \int_{0}^{1}\frac{x^{2}\ln\left(x^{10}+x^{9}+x+1\right)}{1+x^{3}}dx \] We factoriseren de polynoom in de $\ln$ functie: \[x^{10}+x^{9}+x+1=x^9(x+1)+(x+1)=(x+1)(x^9+1)\] Sinds $\ln(ab)=\ln(a)+\ln(b)$: \[\ln\left(x^{10}+x^{9}+x+1\right)=\ln\left(x+1\right)+\ln\left(x^{9}+1\right)\] Door lineariteit geldt: \[ \int_{0}^{1}\frac{x^{2}\ln\left(x^{10}+x^{9}+x+1\right)}{1+x^{3}}dx = \underbrace{\int_{0}^{1}\frac{x^{2}\ln\left(x+1\right)}{1+x^{3}}dx}_{I_1}+\underbrace{\int_{0}^{1}\frac{x^{2}\ln\left(x^{9}+1\right)}{1+x^{3}}dx}_{I_2}\]

Substitutie en partieel primitiveren ($I_2$)

$t=x^3 \implies dt=3x^2dx \implies dx=\frac{dt}{3x^2}$: \[\frac{1}{3}\int_{0}^{1}\frac{\ln\left(t^{3}+1\right)}{1+t}\] $u=\ln(t^3+1) \implies du=\frac{3t^2}{t^3+1} \qquad dv=\frac{1}{t+1} \implies v=\ln(t+1)$: \[\int{udv}=uv-\int{vdu}\] \[\frac{1}{3}\int{\frac{\ln\left(t^{3}+1\right)}{1+t}}=\frac{1}{3}\ln(t^3+1)\ln(t+1)-\frac{1}{3}\int{\frac{3t^2\ln(t+1)}{t^3+1}dt}\] \[\frac{1}{3}\int{\frac{\ln\left(t^{3}+1\right)}{1+t}}=\frac{1}{3}\ln(t^3+1)\ln(t+1)-\int{\frac{t^2\ln(t+1)}{t^3+1}dt}\] Zie dat: \[\int{\frac{t^2\ln(t+1)}{t^3+1}dt}=I_1\] Dus: \[I_2=\frac{1}{3}\ln(t^3+1)\ln(t+1)-I_1\] \[\int_{0}^{1}\frac{x^{2}\ln\left(x^{10}+x^{9}+x+1\right)}{1+x^{3}}dx=I_1+I_2\] \[\implies \int_{0}^{1}\frac{x^{2}\ln\left(x^{10}+x^{9}+x+1\right)=}{1+x^{3}}dx=I_1+\left.\frac{1}{3}\ln(t^3+1)\ln(t+1)\right|^{x=1}_{x=0}-I_1\] \[=\left.\frac{1}{3}\ln(t^3+1)\ln(t+1)\right|^{x=1}_{x=0}\] $t=x^3$: \[\left.\frac{1}{3}\ln(x^9+1)\ln(x^3+1)\right|^{x=1}_{x=0}\] \[=\frac{\ln(2)^2}{3}\]