1∫0x2+2x2+4x+4dx = 1∫0(x+2)2−4(x+2)+6(x+2)2dx = 1∫0(1−4x+2+6(x+2)2)dx = $(x - 4 \ln (x + 2) - \frac{6}{x + 2}) I^{1}_{0}$ = $2 - 4\ln \frac{3}{2}$∣∣∣π4
1∫0x2+2x2+4x+4dx = 1∫0(x+2)2−4(x+2)+6(x+2)2dx = $\int\limits_{0}^{1}$ ($1 - \frac{4}{x + 2} + \frac{6}{(x + 2)^{2}}) dx$ = ($x - 4 \ln (x + 2) - \frac{6}{x + 2}) I^{1}_{0}$ = 2 - 4$\ln\frac{3}{2}$∣∣∣π4