Given: #int(7lnx)/x^3dx#
Rewrite the integral as:
#7intlnx/x^3=>7intlnx*1/x^3dx#
Apply Integration by Parts (I.B.P)
#int# #u# #dv=uv-int# #v# #du#
Let:
#u=lnx=>du=1/xdx#
#dv=1/x^3dx=>v=-1/(2x^2)larr# See proof below
#--------------------#
Use #color(blue)(intx^adx=(x^(a+1))/(a+1)#
#color(red)(int1/x^3dx=intx^(-3)dx=(x^(-3+1))/(-3+1)=x^-2/-2=-1/(2x^2)#
#--------------------#
Substituting into the I.P.B formula:
#=(lnx)(-1/(2x^2))-int(-1/(2x^2))(1/xdx)#
Take out the constant #-1/2#
#=-lnx/(2x^2)--1/2int1/(x^3)dx#
Simplify
#=-lnx/(2x^2)+1/2int1/(x^3)#
Now we solve for #int1/x^3# using which we have already done earlier:
#=-lnx/(2x^2)+1/2[-1/(2x^2)]#
#=-lnx/(2x^2)-1/(4x^2)#
#=7[-lnx/(2x^2)-1/(4x^2)]#
#=-(7lnx)/(2x^2)-7/(4x^2)+"C"larr# Don't forget the constant!
