Question

Integrate :

$\int\limits_{​ }^{​}​( ln(lnx) + \frac{1}{ {ln}^{2}x }) ​\,​ dx$
​

Answer 1

Given Question is

$\displaystyle \int \rm \: \bigg(ln(ln \: x) + \dfrac{1}{ {(ln \: x)}^{2} } \bigg) \: dx$

$\large\underline{\sf{Solution-}}$

Given integral is

$\displaystyle \int \rm \: \bigg(ln(ln \: x) + \dfrac{1}{ {(ln \: x)}^{2} } \bigg) \: dx$

To evaluate this integral, we use method of Substitution.

So, Substitute

$\rm \: ln \: x = y$

$\rm\implies \:x = {e}^{y}$

$\rm\implies \:dx = {e}^{y} \: dy$

So, above integral can be rewritten as

$\rm \:  =  \: \displaystyle \int \rm \: \bigg(ln(y) + \dfrac{1}{ {y}^{2} } \bigg) {e}^{y} \: dy$

can be further rewritten as

$\rm \:  =  \: \displaystyle \int \rm \: \bigg(ln(y) + \frac{1}{y} - \frac{1}{y} + \dfrac{1}{ {y}^{2} } \bigg) {e}^{y} \: dy$

$\rm \:  =  \: \displaystyle \int \rm \: \bigg(ln(y) + \dfrac{1}{y} \bigg) {e}^{y} \: dy \: - \: \displaystyle \int \rm \: \bigg(\dfrac{1}{y} - \dfrac{1}{ {y}^{2} } \bigg) {e}^{y} \: dy$

We know,

$\boxed{\tt{\displaystyle \int \rm \: {e}^{x}(f(x) + f'(x)) \: dx \: = \: {e}^{x}f(x) + c \: }}$

So, using this, we get

$\rm \:  =  \: {e}^{y}ln(y) - {e}^{y}\dfrac{1}{y} + c$

$\rm \:  =  \: {e}^{y}\bigg(ln(y) - \dfrac{1}{y}\bigg) + c$

$\rm \:  =  \: x\bigg(ln(ln \: x) - \dfrac{1}{ {e}^{x} }\bigg) + c$

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Formula Used

$\boxed{\tt{ \dfrac{d}{dx}ln(x) = \frac{1}{x} \: }}$

$\boxed{\tt{ \dfrac{d}{dx} \frac{1}{x} = \frac{ - 1}{ {x}^{2} } \: }}$

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MORE TO KNOW

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Answer:

119

Down vote

Accepted

Put x=tanθ, then your integral transforms to

I=∫π/40log(1+tanθ) dθ.(1)

Now using the property that

∫a0f(x) dx=∫a0f(a−x) dx,

we have

I=∫π/40log(1+tan(π4−θ)) dθ=∫π/40log(21+tanθ) dθ.(2)

Adding (1) and (2) we get

2I=∫π/40log(2) dθ⇒I=log(2)⋅π8.