What is #int_1^(e^(pi/4)) 4/(x(1+(lnx)^2))dx#?

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What is #int_1^(e^(pi/4)) 4/(x(1+(lnx)^2))dx#?

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Answer 1 · 2018-03-16T10:56:12

#4tan^-1(pi/4)#

Explanation:

#I=int_1^(e^(pi/4))4/(x(1+(lnx)^2))dx#
Let, #lnx=u=>1/x*dx=du#
#x=1=>u=ln1=>u=0and#
#x=e^(pi/4)=>u=lne^(pi/4)=>u=pi/4#
#:.I=int_0^(pi/4)4/(1+u^2)du=[4tan^-1u]_0^(pi/4)#
#:.I=4[tan^-1(pi/4)-tan^-1(0)]=4[tan^-1(pi/4)-0]#
#:.I=4tan^-1(pi/4)#

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What is #int_1^(e^(pi/4)) 4/(x(1+(lnx)^2))dx#?

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