Să se calculeze l=limα→∞∫0α2x+1x4+2x3+3x2+2x+2 dxl=\lim_{\alpha\to\infty}\displaystyle\int_0^{\alpha}\dfrac{2x+1}{x^4+2x^3+3x^2+2x+2}\,dxl=limα→∞∫0αx4+2x3+3x2+2x+22x+1dx.
a) l=π3l=\dfrac{\pi}{3}l=3π; b) l=arctg2l=\operatorname{arctg} 2l=arctg2; c) l=π2l=\dfrac{\pi}{2}l=2π; d) l=arctg13l=\operatorname{arctg}\dfrac{1}{3}l=arctg31; e) l=arctg3l=\operatorname{arctg} 3l=arctg3; f) l=π4l=\dfrac{\pi}{4}l=4π.