How do I evaluate the following indefinite integral by using integration by parts? #int(7lnx)/x^3dx#

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#int(7lnx)/x^3dx#

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Answer 1 · 2018-01-19T01:07:42

#int(7lnx)/x^3dx=-(7lnx)/(2x^2)-7/(4x^2)+"C"#

Given: #int(7lnx)/x^3dx#

Rewrite the integral as:

#7intlnx/x^3=>7intlnx*1/x^3dx#

#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!