Note that, 5x3+12=5∗(x3−5x2+4x)+25x2−20x−12
∫5x3+12x3−5x2+4xdx=∫5 dx+∫25x2−20x−12x3−5x2+4x=5x+∫Ax+∫Bx−4+∫Cx−1=5x−3ln|x|+ 773ln|x−4| +73ln|x−1|+ C
∫5x3+12x3−5x2+4xdx=∫5 dx+∫25x2−20x−12x3−5x2+4x=5x+∫Ax+∫Bx−4+∫Cx−1=5x−3ln|x|+ 773ln|x−4| +73ln|x−1|+ C
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