Integraal is te vinden op The Daily Integral 18-12-2025.
\[ \int_{0}^{1}\frac{\ln\left(1+x^{2}\right)}{\left(1+x\right)^{2}}dx \]Partiële Integratie
\[ \int_{0}^{1}\frac{\ln\left(1+x^{2}\right)}{\left(1+x\right)^{2}}dx \] \[ u=\ln(1+x^2), du=\frac{2x}{1+x^2}, dv=(1+x)^-2, v=-(1+x)^-1 \] \[ -\ln\left(1+1^{2}\right)\left(1+1\right)^{-1}+\ln\left(1+0^{2}\right)\left(1+0\right)^{-1} -\int_{0}^{1}-\left(1+x\right)^{-1}\frac{2x}{1+x^{2}}dx \] \[ - \frac{\ln\left(2\right)}{2} +2\int_{0}^{1}\frac{x}{(1+x^{2})(x+1)}dx \]Breuken Splitsen
\[ - \frac{\ln\left(2\right)}{2} +2\int_{0}^{1}\frac{x}{(1+x^{2})(x+1)}dx \]Dit is een breuk met één lineaire en één kwadratische term in de noemer, dus hebben de 3 graden van beweging nodig: \(A, B, C\).
\[ \left(Ax+B\right)\left(1+x\right)+C\left(1+x^{2}\right)=x \] \[ Ax+B+Ax^{2}+Bx+Cx^{2}+C=x \] \[ \left(A+C\right)x^{2}+\left(A+B\right)x+\left(B+C\right)=x \] \[A+C=0\] \[A+B=1\] \[B+C=0\] \[A=B= \frac{1}{2}, C= -\frac{1}{2} \] \[ - \frac{\ln\left(2\right)}{2} +2\int_{0}^{1}\left(\frac{\frac{1}{2}x+\frac{1}{2}}{1+x^{2}}+\frac{-\frac{1}{2}}{1+x}\right)dx \] \[ - \frac{\ln\left(2\right)}{2} + \int_{0}^{1}\frac{x}{x^{2}+1}dx+\int_{0}^{1}\frac{1}{x^{2}+1}dx-\int_{0}^{1}\frac{1}{x+1}dx \]\(\arctan\)-integraal
De afgeleide van \( \arctan(x) \) is \( \frac{1}{x^2+1} \):
\[ - \frac{\ln\left(2\right)}{2} +\int_{0}^{1}\frac{x}{x^{2}+1}dx + \arctan(1)-\arctan(0) -\int_{0}^{1}\frac{1}{x+1}dx \]\( \ln \)-integraal
\[ - \frac{\ln\left(2\right)}{2} +\int_{0}^{1}\frac{x}{x^{2}+1}dx+ \arctan(1)-\arctan(0) -\int_{0}^{1}\frac{1}{x+1}dx \] \[u=x+1, du=1dx, dx=du \] \[ - \frac{\ln\left(2\right)}{2} +\int_{0}^{1}\frac{x}{x^{2}+1}dx+ \arctan(1)-\arctan(0) -\int_{1}^{2}\frac{1}{u}du \]De afgeleide van \( \ln(x) \) is \( \frac{1}{x} \):
\[ - \frac{\ln\left(2\right)}{2} +\int_{0}^{1}\frac{x}{x^{2}+1}dx+ \arctan(1)-\arctan(0) -\ln(2)+\ln(1) \]U substitutie
\[ - \frac{\ln\left(2\right)}{2} +\int_{0}^{1}\frac{x}{x^{2}+1}dx+ \arctan(1)-\arctan(0) -\ln(2)+\ln(1) \] \[u=x^2+1, du=2xdx, dx=\frac{du}{2x} \] \[ - \frac{\ln\left(2\right)}{2} +\frac{1}{2} \int_{1}^{2}\frac{1}{u}du+ \arctan(1)-\arctan(0) -\ln(2)+\ln(1) \] \[ - \frac{\ln\left(2\right)}{2} +\frac{1}{2}\ln(2)-\frac{1}{2}\ln(1) + \arctan(1)-\arctan(0) -\ln(2)+\ln(1) \]Herleiden
\[ - \frac{\ln\left(2\right)}{2} +\frac{1}{2}\ln(2)-\frac{1}{2}\ln(1) + \arctan(1)-\arctan(0) -\ln(2)+\ln(1) \] \[= \arctan(1)-\ln(2) \]Exacte waarde \(\arctan\)
\[ \arctan(1)-\ln(2) \] \[ \tan(x)=1 \] \[ \frac{\sin(x)}{\cos(x)} = 1 \] \[ \sin(x) = \cos(x) \]Sinus en Cosinus zijn gelijk op 45 graden (\( \frac{\pi}{4} \) rad):
\[ \arctan(1) = \frac{\pi}{4} \]