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ax-b-cx-d-2-dx-

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Question Number 5032 by gourav~ last updated on 04/Apr/16
∫((ax+b)/((cx+d)^2 ))dx =?
∫ax+b(cx+d)2dx=?
Answered by Yozzii last updated on 04/Apr/16
Let u=cx+d⇒du=cdx⇒c^(−1) du=dx. x=((u−d)/c)⇒ax+b=((a(u−d))/c)+b ax+b=ac^(−1) u−adc^(−1) +b ∫((ax+b)/((cx+d)^2 ))dx=∫(((ac^(−1) u−adc^(−1) +b)c^(−1) du)/u^2 ) =c^(−1) ∫(ac^(−1) u^(−1) +(b−adc^(−1) )u^(−2) )du =c^(−1) (ac^(−1) ln∣u∣+(adc^(−1) −b)u^(−1) )+K ∫((ax+b)/((cx+d)^2 ))dx=((aln∣cx+d∣)/c^2 )+((ad−bc)/(c^2 (cx+d)))+K (c≠0) K=constant of integration
Letu=cx+d⇒du=cdx⇒c−1du=dx.x=u−dc⇒ax+b=a(u−d)c+bax+b=ac−1u−adc−1+b∫ax+b(cx+d)2dx=∫(ac−1u−adc−1+b)c−1duu2=c−1∫(ac−1u−1+(b−adc−1)u−2)du=c−1(ac−1ln∣u∣+(adc−1−b)u−1)+K∫ax+b(cx+d)2dx=aln∣cx+d∣c2+ad−bcc2(cx+d)+K(c≠0)K=constantofintegration

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