Pour [tex] \int\limits^x_1 {log(1+ t^{2}})\, dt [/tex] on peut faire une intégration par parties :
[tex]\int\limits^x_1{log(1+t^{2}})\,dt=[t*log(1+t^{2})]-\int\limits^x_1{t*\frac{2t}{1+t^{2}}}\,dt [/tex]
Or t²/(1+t²)=1-1/(1+t²)
Donc [tex]\int\limits^x_1{\frac{t^{2}}{1+t^{2}}}\,dt=\int\limits^x_1{1}\,dt-\int\limits^x_1{\frac{1}{1+t^{2}}}\,dt [/tex]
Or une primitive de 1/(1+x²) est arctanx
Donc [tex] \int\limits^x_1 {\frac{t^{2}}{1+t^{2}}}\,dt=x-1-[arctant]=x-1-arctanx+arctan1[/tex]
Donc [tex]\int\limits^x_1{log(1+t^{2}})\,dt=x*log(1+x^{2})-log(2)-2*(x-arctanx-1+\frac{\pi}{4}) [/tex]
Donc f(x)=x²*log(1+x²)-x*log(2)-2x²+2x*arctanx+2x-πx/2
f'(x)=2xlog(1+x²)+2x³/(1+x²)-log(2)-4x+2arctanx+2x/(1+x²)+2-π/2
f'(x)=2xlog(1+x²)+(2x³+2x)/(1+x²)-4x+2arctanx-log(2)+2-π/2
