How do I find the integral #int10/((x-1)(x^2+9))dx# ?

#=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