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Question Number 61855 by aliesam last updated on 10/Jun/19
∫013x3−x2+2x−4x2−3x+2dx
Answered by MJS last updated on 10/Jun/19
3x3−x2+2x−4x2−3x+2=−(3x2+2x+4)x−1x−2forx<2−∫(3x2+2x+4)x−1x−2dx=[t=x−1x−2→dx=−2(x−2)3(x−1)]=∫2t2(20t4−26t2+9)(t2−1)4dt==∫(38(t−1)4+72(t−1)3+18516(t−1)2+13516(t−1)+38(t+1)4−72(t+1)3+18516(t+1)2−13516(t+1))dt==−18(t−1)3−74(t−1)2−18516(t−1)+13516ln(t−1)−18(t+1)3+74(t+1)2−18516(t+1)−13516ln(t+1)==−t(185t4−312t2+135)8(t2−1)3+13516lnt−1t+1==−18(8x2+26x+101)x2−3x+2+13516ln(2x−3−2x2−3x+2)+C∫013x3−x2+2x−4x2−3x+2dx=−10128−13516ln(3−22)
Commented by aliesam last updated on 10/Jun/19
thankssirbrilliantsolution
Commented by MJS last updated on 10/Jun/19
you′rewelcome
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