Evaluate : ∫x2+5x+6x+2dxAnswer: x2+5x+6−21log(x+25)+x2+5x+6+CSolutionWrite x+2=21(2x+5)−21, where 2x+5=dxd(x2+5x+6). Then∫x2+5x+6x+2dx=21∫x2+5x+6(2x+5)dx−21∫x2+5x+6dx.The first integral is x2+5x+6. For the second, x2+5x+6=(x+25)2−(21)2:−21log(x+25)+x2+5x+6.x2+5x+6−21log(x+25)+x2+5x+6+CCBSE Class 12 Board · 2014 · Subjective