How do I solve the integral using partial fractions?

Question

#int4/((x+a)(x+b))dx#

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Answer 1 · 2017-09-19T07:53:09

#4/(b-a)ln|(x+a)/(x+b)|+C" "a!=b#

#int4/((x+a)(x+b))dx#

consider the integrand

#4/((x+a)(x+b))-=A/(x+a)+B/(x+b)#

multiply both sides by #(x+a)(x+b)#

this gives on simplification

#4-=A(x+b)+B(x+a)--(1)#

#x=-a" into "(1)#

#4=A(b-a)+0#

#:.A=4/(b-a), a!=b#

#x=-b" into "(1)#

#4=0+B(a-b)#

#B=4/(a-b)=-4/(b-a),a!=b#

#:.4/((x+a)(x+b))-=4/((b-a)(x+a))-4/((b-a)(x+b)#

#int4/((x+a)(x+b))=int[4/((b-a)(x+a))-4/((b-a)(x+b)]]dx#

#=4/(b-a)int[1/(x+a)-1/(x+b)]dx#

using

#color(blue)(int(f'(x))/f(x)dx=ln|f(x)|+C)#

#=4/(b-a)[ln|(x+a)|-ln|(x+b)|]+C#

#=4/(b-a)ln|(x+a)/(x+b)|+C" "a!=b#