How do I find the integral #int10/((x-1)(x^2+9))dx# ?
1 Answer
#=5/4ln(x-1)-5/8ln(x^2+9)+5/12tan^-1(x/3)+c# , where#c# is a constant
Explanation :
This type of question usually solve by using Partial Fractions,
#10/((x-1)(x^2+9))# , it can be written as
#10/((x-1)(x^2+9))=A/(x-1)+(Bx+C)/(x^2+9)# multiplying both sides by
#(x-1)(x^2+9)# , we get
#10=A(x^2+9)+(Bx+C)(x-1)#
#10=(A+B)x^2+(C-B)x+(9A-C)# comparing coefficients of
#x^2# ,#x# and constant both side,
#A+B=0# #=># #A=-B# .................#(i)#
#C-B=0# #=># #C=-B# .................#(ii)#
#9A-C=10# #=># .................#(iii)# substituting
#A# &#C# from#(i)# and#(ii)# in#(iii)#
#9(-B)-(-B)=10#
#B=-10/8=-5/4# Therefore,
#A=C=5/4# Hence, we get
#10/((x-1)(x^2+9))=5/(4(x-1))+((-5/4)x+(5/4))/(x^2+9)# integrating both sides with respect to
#x# ,
#int10/((x-1)(x^2+9))dx=int5/(4(x-1))dx+int((-5/4)x+(5/4))/(x^2+9)dx#
#=5/4int1/((x-1))dx-5/4int(x)/(x^2+9)dx+5/4int1/(x^2+9)dx# Second integral is solved by substitution, as the numerator is the differentiation of denominator
#=5/4ln(x-1)-5/4*1/2ln(x^2+9)+5/4*1/3tan^-1(x/3)#
#=5/4ln(x-1)-5/8ln(x^2+9)+5/12tan^-1(x/3)+c# , where#c# is a constant
