How do you evaluate #\int \frac { 2x d x } { ( x + 1) ( x + 3) }#?

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Answer 1 · 2017-12-02T10:29:30

#3ln|(x+3)|-ln|(x+1)|+C#

we will need to split the integrand into partial fractions

#I=int(2x)/((x+1)(x+3))dx--(1)#

so

#(2x)/((x+1)(x+3))-=A/(x+1)+B/(x+3)#

multiply both sides by #(x+1)(x+3)# and cancel

#2x-=A(x+3)+B(x+1)--(2)#

#x=-1#

#(2)rarr-2=A(-1+3)+0#

#-2=2A=>A=-1#

#x=-3#

#(2)rarr2xx-3=0+B(-3+1)#

#-6=-2B=>B=3#

#(1)I=rarrint(-1/(x+1)+3/(x+3))dx#

using

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

#:.I=-ln|(x+1)+3ln|(x+3)|+C#

#=3ln|(x+3)|-ln|(x+1)|+C#