Integrate : x/x3-1dx Share with your friends Share 0 Manbar Singh answered this Let I = ∫x dxx3-1 = ∫x dxx-1x2+x+1Let xx-1x2+x+1 = Ax-1 + Bx + Cx2+x+1⇒x = Ax2+x+1 + Bx+Cx-1⇒x = A+Bx2 + A-B+Cx + A-CComparing the coefficient of x2 on both sides,A + B = 0⇒A = -B .....1Comparing the coefficient of x on both sides,A-B+C=1 .....2Comparing constant term on both sides, A-C = 0⇒A = C ....3From 1 and 3, A=C=-BNow, from 2, we get-B-B-B=1⇒B= -13Now, A=C=13So, xx-1x2+x+1 = 1/3x-1 + -1/3x + 1/3x2+x+1xx-1x2+x+1=13×1x-1-13×xx2+x+1 + 13×1x2+x+1∫x dxx-1x2+x+1=13∫dxx-1-13∫xdxx2+x+1 + 13∫dxx2+x+1=13logx-1 - 16∫2x+1-1x2+x+1 dx + 13∫dxx2+x+1=13logx-1 - 16∫2x+1x2+x+1 dx + 16∫dxx2+x+1+13∫dxx2+x+1I=13logx-1 - 16 ∫2x+1x2+x+1 dx + 12∫dxx2+x+1Let I1 =∫2x+1x2+x+1 dxput x2+x+1 = t⇒2x+1dx = dtso, I1 = ∫dtt = log t = logx2+x+1So, I=13logx-1 -16logx2+x+1+12∫dxx2+x+14+34I=13logx-1 -16logx2+x+1+12∫dxx+122+322I=13logx-1 -16logx2+x+1+12×23tan-1x+123/2 + CI=13logx-1 -16logx2+x+1+13tan-12x+13+C 1 View Full Answer
