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Question Number 113675 by AbhishekBasnet last updated on 14/Sep/20

Commented by mohammad17 last updated on 14/Sep/20

=− (1/b)∫((a−bx)−a)(a−bx)^(−(1/2)) )dx =−(1/b)∫((a−bx)^(1/2) −a(a−bx)^(−(1/2)) )dx =(2/(3b^2 ))(√((a−bx)^3 )) − ((2a)/b^2 )(√(a−bx))+C by:⟨miss.moh⟩(method 1)

=−1b∫((a−bx)−a)(a−bx)−12)dx=−1b∫((a−bx)12−a(a−bx)−12)dx=23b2(a−bx)3−2ab2a−bx+Cby:⟨miss.moh⟩(method1)

Commented by mohammad17 last updated on 14/Sep/20

method (2) by ⟨miss.moh⟩

method(2)by⟨miss.moh⟩

Commented by mohammad17 last updated on 14/Sep/20

by parts: u=x⇒u^′ =dx ,v^′ =∫ (dx/( (√(a−bx))))⇒v=−(2/b)(√(a−bx)) =−((2x)/b)(√(a−bx))+(2/b)∫ (√(a−bx))dx =−((2x)/b)(√(a−bx)) −(4/(3b^2 ))(√((a−bx)^3 )) +C

byparts:u=x⇒u′=dx,v′=∫dxa−bx⇒v=−2ba−bx=−2xba−bx+2b∫a−bxdx=−2xba−bx−43b2(a−bx)3+C

Answered by Dwaipayan Shikari last updated on 14/Sep/20

∫(x/( (√(a−bx)))) dx a−bx=t^2 ,−b=2t(dt/dx) −(1/b)∫(((a−t^2 )/b)/t)2tdt (2/b^2 )∫t^2 −a dt (2/(3b^2 ))t^3 −((2at)/b^2 )+C=(2/(3b^2 ))(a−bx)^(3/2) −((2a)/b)(√(a−bx)) +C

∫xa−bxdxa−bx=t2,−b=2tdtdx−1b∫a−t2bt2tdt2b2∫t2−adt23b2t3−2atb2+C=23b2(a−bx)32−2aba−bx+C

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