Here,
#I=int(x+1)/((x^2+4)(3x-5))dx#
To obtain Partial Fractions:
#color(blue)((x+1)/((x^2+4)(3x-5))=A/(3x-5)+(Bx+C)/(x^2+4)#
#=>x+1=A(x^2+4)+B(3x^2-5x)+C(3x-5)#
#=>x+1=x^2(A+3B)+x(-5B+3C)+(4A-5C)#
comparing coefficients of #x^2 ,x and #constant term :
#A+3B=0 =>A=-3Bto(1) ,#
#-5B+3C=1=>3C=1+5B=>C=(1+5B)/3to(2)# ,
#4A-5C=1to(3)#
Putting value of #A and C# into #(3)#
#4(-3B)-5((1+5B)/3)=1=>-36B-5-25B=3#
#:. color(blue)(B=-8/61#
From #(1)# ,we get , #A=-3(-8/61)=>color(blue)(A=24/61#
From #(2) #, we get #C=(1+5(-8/61))/3=(61-40)/(3*61)=>color(blue)(C=7/61#
So,
#I=1/61int{24/(3x-5)+(-8x+7)/(x^2+4)}dx#
#=1/61{8int3/(3x-5)dx-4int(2x)/(x^2+4)dx+7int1/(x^2+2^2)dx}#
#=1/61{8ln|3x-5|-4ln|x^2+4|+7*1/2 tan^-1(x/2)}+c#
Hence,
#I=1/61{8ln|3x-5|-4ln|x^2+4|+7/2 tan^-1(x/2)}+c#
