How do you find #\int _ { 2} ^ { 4} \frac { 1} { x ( \ln x + 3) ^ { 2} } d x#?

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Answer 1 · 2018-02-17T11:56:24

#=1/(ln2+3)-1/(ln4+3)#

#int_2^4(1/(x(lnx+3)^2))dx#

now

#d/(dx)(1/(lnx+3))#

using the quotient rule gives us

#(dy)/(dx)=((0-(1/x))/(lnx+3)^2)#

#=-1/(x(lnx+3)^2)#

which means we can do the integral by inspection

#int(1/(x(lnx+3)^2))dx=-[1/(lnx+3)]_2^4#

#-[1/(ln4+3)-1/(ln2+3)]#

#=1/(ln2+3)-1/(ln4+3)#

alternatively we can use substitution

#int_2^4(1/(x(lnx+3)^2))dx#

#u=lnx+3=>du=1/xdx=>dx=xdu#

omitting the limits for now

#int(1/(x(lnx+3)^2))dx=int(1/(x*u^2))xdu#

#=int1/u^2dx=-1/u#

#

#=substitute back for u and substitute the limits as before

#-[1/(lnx+3)]_2^4#