Question

Find $\int {\left[ {\log \left( {\log x} \right) + \dfrac{1}{{{{\left( {\log x} \right)}^2}}}} \right]dx}$.

Answer 1

Hint:
We can expand the integral to both the terms by opening the bracket. Then we can find the integral of the $1^{\text{st}}$ term by the method of integration by parts. Then we can integrate the $2^{\text{nd}}$ term after integration by parts again using the same method. Then we can simplify the expression.

Complete step by step solution:
We need to find the integral, $\int {\left[ {\log \left( {\log x} \right) + \dfrac{1}{{{{\left( {\log x} \right)}^2}}}} \right]dx} $ ,
Let $I = \int {\left[ {\log \left( {\log x} \right) + \dfrac{1}{{{{\left( {\log x} \right)}^2}}}} \right]dx} $
We can open the bracket. So, we get,
 \[ \Rightarrow I = \int {\log \left( {\log x} \right)dx} + \int {\dfrac{1}{{{{\left( {\log x} \right)}^2}}}dx} \]
Let \[{I_1} = \int {\log \left( {\log x} \right)dx} \]
We can find the integral by using the method of integration by parts.
We know that $\int {uvdx} = u\int {vdx} - \int {\left( {u'\int {vdx} } \right)dx} $ .
Here we take \[u = \log \left( {\log x} \right)\] and $v = 1$
Now the derivative of u is given by,
 \[ \Rightarrow u' = \dfrac{d}{{dx}}\log \left( {\log x} \right)\]
We know that $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$ and by applying chain rule, we get,
 \[ \Rightarrow u' = \dfrac{1}{{\log x}} \times \dfrac{d}{{dx}}\log x\]
Further using same property, we get,
 \[ \Rightarrow u' = \dfrac{1}{{x\log x}}\] …. (1)
The integral of v is given by,

$ \Rightarrow \int {vdx} = \int {dx} $
As \[\int {dx = x} \] , we get,
 $ \Rightarrow \int {vdx} = x$ .. (2)
Now we can write \[{I_1}\] using integration by parts and applying equations (1) and (2)
 \[ \Rightarrow {I_1} = \log \left( {\log x} \right)x - \int {\left( {\dfrac{1}{{x\log x}}x} \right)dx} \]
On simplification, we get,
 \[ \Rightarrow {I_1} = \log \left( {\log x} \right)x - \int {\dfrac{1}{{\log x}}dx} \]
Now we can again simplify the integral using integration by parts. Here \[u = \dfrac{1}{{\log x}}\] and $v = 1$ .
 \[ \Rightarrow {I_1} = \log \left( {\log x} \right)x - \left[ {\dfrac{1}{{\log x}}\int {1dx - } \int {\dfrac{d}{{dx}}\left( {\dfrac{1}{{\log x}}} \right)\int {1dx} \,dx} } \right]\]
We know that $\int {1dx} = x$ and \[\dfrac{d}{{dx}}\left( {\dfrac{1}{{\log x}}} \right) = - \dfrac{1}{{{{\left( {\log x} \right)}^2}}} \times \dfrac{d}{{dx}}\log x = - \dfrac{1}{{x{{\left( {\log x} \right)}^2}}}\]
 \[ \Rightarrow {I_1} = \log \left( {\log x} \right)x - \left[ {\dfrac{x}{{\log x}} - \int {\left( { - \dfrac{1}{{x{{\left( {\log x} \right)}^2}}}} \right)x\,dx} } \right]\]
On cancelling the common terms, we get,
 \[ \Rightarrow {I_1} = \log \left( {\log x} \right)x - \left[ {\dfrac{x}{{\log x}} + \int {\dfrac{1}{{{{\left( {\log x} \right)}^2}}}\,dx} } \right]\]
On opening the square bracket, we get,
 \[ \Rightarrow {I_1} = \log \left( {\log x} \right)x - \dfrac{x}{{\log x}} - \int {\dfrac{1}{{{{\left( {\log x} \right)}^2}}}\,dx} \]
Now we can substitute in I. So, we get,
 \[ \Rightarrow I = \log \left( {\log x} \right)x - \dfrac{x}{{\log x}} - \int {\dfrac{1}{{{{\left( {\log x} \right)}^2}}}\,dx} + \int {\dfrac{1}{{{{\left( {\log x} \right)}^2}}}dx} \]
On simplification, we get,
 \[ \Rightarrow I = x\log \left( {\log x} \right) - \dfrac{x}{{\log x}}\]
On adding the constant of integration, we get,
 \[ \Rightarrow I = x\log \left( {\log x} \right) - \dfrac{x}{{\log x}} + C\]

So, the required integral is \[x\log \left( {\log x} \right) - \dfrac{x}{{\log x}} + C\].

Note:
We cannot integrate both the terms to find the integral as we cannot find the integral for the $2^{\text{nd}}$ term. While using the method of integration by parts, we must take the $2^{\text{nd}}$ function as 1 so that we can calculate its integral as x. The order of taking the functions for integrating by parts is given by inverse trigonometric function, logarithmic function, arithmetic function, trigonometric function and exponential function. While expanding a negative term, we must use brackets properly to avoid the errors.