What is $\int \frac{1 + ln x}{x^{2}} d x$ $int (1+lnx)/x^2dx$ ?

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Answer 1 · 2018-03-15T01:42:11

$int \ (1+lnx)/x^2 \ dx = -(2+lnx)/x C$

We seek:

$I = int \ (1+lnx)/x^2 \ dx$

In preparation for an application of [integration by Parts]We can then apply Integration By Parts:

Let ${ (u,=1+lnx, => (du)/dx,=1/x), ((dv)/dx,=1/x^2, => v,=-1/x ) :}$

Then plugging into the IBP formula:

$int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx$

We have:

$int \ (1+lnx)(1/x^2) \ dx = (1+lnx)(-1/x) - int \ (-1/x)(1/x) \ dx$

Giving is:

$I = -(1+lnx)/x + int \ 1/x^2 \ dx$

$\ \ = -(1+lnx)/x -1/x + C$

$\ \ = -(2+lnx)/x C$

What is int (1+lnx)/x^2dx∫1+lnxx2dxint (1+lnx)/x^2dx?