What is the antiderivative of $(ln x) ^ 2 / x ^ 2$ ?

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Answer 1 · 2016-07-01T12:36:50

$I=-(ln x)^2/x-(2ln x)/x-2/x+c$

The given function is

$(ln x) ^ 2 / x ^ 2$

We are to find out $I = int (ln x) ^ 2 / x ^ 2dx$

Let
$ln x =u =>x=e^u$

Differentiating w.r.t x we have

$(d(x))/(dx) =(d(e^u))/(du)*(du)/(dx)$

$=>1 =e^u*(du)/(dx)$

$=>dx=e^udu$

Now changing the variable the integral becomes

$I=intu^2/(e^(2u))*e^udu=intu^2*e^-udu$

$I= u^2inte^-udu-int((d(u^2))/(du)inte^-udu)du$

$= u^2inte^-udu-int2u(-e^-u)du$

$= u^2inte^-udu+int2ue^-udu$

$= u^2inte^-udu+2intue^-udu$

$= u^2inte^-udu+2uinte^-udu-2int((du)/(du)inte^-udu)du$

$= u^2inte^-udu+2uinte^-udu-2int(-e^-u)du$

$= u^2inte^-udu+2uinte^-udu+2int(e^-u)du$

$=-u^2e^-u-2ue^-u-2e^-u+c$

Inserting $u = lnx and e^-u =1/x$

$I=-(ln x)^2/x-(2ln x)/x-2/x+c$

What is the antiderivative of (ln x) ^ 2 / x ^ 2(ln x) ^ 2 / x ^ 2?