(1)
∫(x3−x2)(3x2−x+1)dx=x3−x23x2−2x+x−1+2dx=x3−x23x2−2xdx+x2(x−1)x−1dx+x2(x−1)2dx
Let x3−x2=t
⇒(3x2−2x)dx=dt and using partial fraction in the third term-
=tdt+x21dx+x−12dx−x2dx−x22dx=lnt−x1+2ln(x−1)−2lnx+x2+C=ln(x3−x2)+x1+2lnxx−1+C
(2)
∫(t+1)3(t2)dt=(t+1)3(t+1−1)2dt=(t+1)3(t+1)2+1−2(t+1)dt=t+11dt+(t+1)31dt−(t+1)22dt=ln(t+1)−2(t+1)21+(t+1)2+C
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