Question

Evaluate the given function $\int {\log \left( {\log x} \right)dx + \int {{{\left( {\log x} \right)}^{ - 2}}dx} } $
A.$x\log \left( {\log x} \right) + c$
B.\[x\log \left( {\log x} \right) + \dfrac{x}{{\log x}} + c\]
C.$x\log \left( {\log x} \right) - \dfrac{x}{{\log x}} + c$
D.$\log x + x\log x + c$

Answer 1

Hint: We will consider this in two part $I = {I_1} + {I_2}$ where ${I_1} = \int {\log y \times {e^y}dy} $ and \[{I_2} = \int {{y^{ - 2}}{e^y}dy} \]. For ${I_1}$we will use integration by part $\int {UVdx = U\int {Vdx - \int {\dfrac{{dU}}{{dx}}\int {Vdxdx} } } } $ where $U = \log y$ and $V = {e^y}$on solving we can eliminate ${I_2}$ and get the value of $I$.

Complete step-by-step answer:
$\int {\log \left( {\log x} \right)dx + \int {{{\left( {\log x} \right)}^{ - 2}}dx} } $
Substitute $\log x = y$, therefore $x = {e^y}$and $dx = {e^y}dy$
$I \Rightarrow \int {\log y \times {e^y}dy + \int {{y^{ - 2}}{e^y}dy} } $
Let ${I_1} = \int {\log y \times {e^y}dy} $ and \[{I_2} = \int {{y^{ - 2}}{e^y}dy} \]
For ${I_1}$ we will use integration by part
 $ \Rightarrow {I_1} = \log y\int {{e^y}dy - \int {\dfrac{{d\log y}}{{dy}}\left( {\int {{e^y}dy} } \right)} } dy$
$ \Rightarrow {I_1} = {e^y}\log y - \int {\dfrac{1}{y}{e^y}dy} $
Again, using integration by part for $\int {\dfrac{1}{y}{e^y}dy} $we get
$ \Rightarrow {I_1} = {e^y}\log y - \dfrac{1}{y}\int {{e^y}dy + \int {\dfrac{{d\dfrac{1}{y}}}{{dy}}\left( {\int {{e^y}dy} } \right)} } dy$
$ \Rightarrow {I_1} = {e^y}\log y - \dfrac{{{e^y}}}{y} - \int {{y^{ - 2}}{e^y}dy} $
$\int {{y^{ - 2}}{e^y}dy} = {I_2}$ assumed previously
So, $ \Rightarrow {I_1} = {e^y}\log y - \dfrac{{{e^y}}}{y} - {I_2} + c$
Now, $I = {I_1} + {I_2}$
Substituting the value of ${I_1}$
$I = {e^y}\log y - \dfrac{{{e^y}}}{y} - {I_2} + {I_2}$
$I = {e^y}\log y - \dfrac{{{e^y}}}{y} + c$
Substituting $x = {e^y}$ and $\log x = y$
$I = {e^{\log x}}\log \left( {\log x} \right) - \dfrac{{{e^{\log x}}}}{{\log x}} + c$
$\therefore I = x\log \left( {\log x} \right) - \dfrac{x}{{\log x}} + c$
Option C is answer

Note: In integration by part $\int {UVdx = U\int {Vdx - \int {\dfrac{{dU}}{{dx}}\int {Vdxdx} } } } $ U and V decided according to ILATE rule
I=inverse trigonometric
L=logarithmic
A=algebra
T=trigonometric
E=exponential
According to ILATE the function which came 1st is said to be U and the next one is V. as seen in the above question ${I_1} = \int {\log y \times {e^y}dy} $,$\log y$ is logarithmic function and ${e^y}$ is exponential, according to ILATE, L came 1st so $\log y$ is consider as U whereas ${e^y}$ consider as V.