What is $\int_{1}^{e^{\frac{π}{4}}} \frac{4}{x (1 + {(ln x)}^{2})} d x$ $int_1^(e^(pi/4)) 4/(x(1+(lnx)^2))dx$ ?

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What is $\int_{1}^{e^{\frac{π}{4}}} \frac{4}{x (1 + {(ln x)}^{2})} d x$ $int_1^(e^(pi/4)) 4/(x(1+(lnx)^2))dx$ ?

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Answer 1 · 2018-03-16T10:56:12

$4 {tan}^{- 1} (\frac{π}{4})$ $4tan^-1(pi/4)$

Explanation:

$I = \int_{1}^{e^{\frac{π}{4}}} \frac{4}{x (1 + {(ln x)}^{2})} d x$ $I=int_1^(e^(pi/4))4/(x(1+(lnx)^2))dx$
Let, $ln x = u \Rightarrow \frac{1}{x} \cdot d x = d u$ $lnx=u=>1/x*dx=du$
$x = 1 \Rightarrow u = ln 1 \Rightarrow u = 0 and$ $x=1=>u=ln1=>u=0and$
$x = e^{\frac{π}{4}} \Rightarrow u = {ln e}^{\frac{π}{4}} \Rightarrow u = \frac{π}{4}$ $x=e^(pi/4)=>u=lne^(pi/4)=>u=pi/4$
$:.I=int_0^(pi/4)4/(1+u^2)du=[4tan^-1u]_0^(pi/4)$
$:.I=4[tan^-1(pi/4)-tan^-1(0)]=4[tan^-1(pi/4)-0]$
$:.I=4tan^-1(pi/4)$

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What is $\int_{1}^{e^{\frac{π}{4}}} \frac{4}{x (1 + {(ln x)}^{2})} d x$ $int_1^(e^(pi/4)) 4/(x(1+(lnx)^2))dx$ ?

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