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

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

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Answer 1 · 2018-05-24T09:46:45

The answer is $= - \frac{5}{x} - 4 \frac{ln (| x |)}{x} - \frac{{(ln (| x |))}^{2}}{x} + C$ $=-5/x-4ln(|x|)/x-(ln(|x|))^2/x+C$

Explanation:

The integral is

$I = \int \frac{{(1 + ln x)}^{2} d x}{x^{2}}$ $I=int((1+lnx)^2dx)/x^2$

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

$= \int \frac{d x}{x^{2}} + \int \frac{2 ln x d x}{x^{2}} + \int \frac{{(ln x)}^{2} d x}{x^{2}}$ $=intdx/x^2+int(2lnxdx)/x^2+int((lnx)^2dx)/x^2$

$= I_{1} + I_{2} + I_{3}$ $=I_1+I_2+I_3$

Therefore,

$I_{1} = \int \frac{d x}{x^{2}} = \int x^{- 2} d x = - \frac{1}{x}$ $I_1=intdx/x^2=intx^-2dx=-1/x$

For the calculation of $I_{2}$ $I_2$ , perform an integration by parts

$intuv'=uv-intu'v$

$u=lnx$ , $=>$ , $u'=1/x$

$v'=2/x^2$ , $=>$ , $v=-2/x$

So,

$I_2=-2lnx/x-2int(-dx)/x^2$

$=-2lnx/x-2/x$

For the calculation of $I_3$ , perform an integration by parts

$u=(lnx)^2$ , $=>$ , $u'=2lnx/x$

$v'=1/x^2$ , $=>$ , $v=-1/x$

$I_3=-(lnx)^2/x-2int(-lnxdx)/x^2$

$=-(lnx)^2/x-2lnx/x-2/x$

Finally,

$I=I_1+I_2+I_3$

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

$=-5/x-4lnx/x-(lnx)^2/x+C$

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

1 Answer

Explanation:

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