How do you evaluate the integral $int 1/(x(1+(lnx)^2)$ ?

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Answer 1 · 2017-02-05T19:52:24

$int1/(x(1+(lnx)^2))dx=arctan(lnx)+C$

Let $u=lnx$ . This implies that $du=1/xdx$ .

$int1/(x(1+(lnx)^2))dx=int1/(1+(lnx)^2)(1/xdx)=int1/(1+u^2)du$

This is the arctangent integral:

$=arctan(u)+C=arctan(lnx)+C$

How do you evaluate the integral int 1/(x(1+(lnx)^2)int 1/(x(1+(lnx)^2)?