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Question: How do you integrate \[ - \dfrac{1}{{x{{\left( {\ln x} \right)}^2}}}\]?...

How do you integrate −1x(ln⁡x)2 - \dfrac{1}{{x{{\left( {\ln x} \right)}^2}}}?

Explanation

Solution

Integration is the process of finding the antiderivative. For, example: The integration of f′(x)f'\left( x \right)with respect to dx is given by∫f′(x)dx=f(x)+C\int {f'\left( x \right)dx = f\left( x \right) + C} , where C is the constant of integration and in the given function, we can find the integral by simplifying the terms by finding its derivative and apply the rule ∫undu=un+1n+1+C\int {{u^n}du = \dfrac{{{u^{n + 1}}}}{{n + 1}} + C} and find the integration of terms.

Complete step by step solution:
The given function is
−1x(ln⁡x)2- \dfrac{1}{{x{{\left( {\ln x} \right)}^2}}}
As given,
∫−1x(ln⁡x)2dx\int { - \dfrac{1}{{x{{\left( {\ln x} \right)}^2}}}dx}
\Rightarrow $$$$\int { - \dfrac{1}{{x{{\left( {\ln x} \right)}^2}}}dx} = - \int {\dfrac{{{{\left( {\ln x} \right)}^{ - 2}}}}{x}dx}
We can use substitution here, since the derivative of ln⁡x\ln x, which is 1x\dfrac{1}{x}, is present alongside ln⁡x\ln x.
Let, u=ln⁡xu = \ln x such that,
du=1xdxdu = \dfrac{1}{x}dx
Hence, we get:
−∫(ln⁡x)−2xdx=−∫(ln⁡x)−2(1x)dx- \int {\dfrac{{{{\left( {\ln x} \right)}^{ - 2}}}}{x}dx} = - \int {{{\left( {\ln x} \right)}^{ - 2}}\left( {\dfrac{1}{x}} \right)dx}
\Rightarrow $$$$ - \int {\dfrac{{{{\left( {\ln x} \right)}^{ - 2}}}}{x}dx} = - \int {{{\left( {\ln x} \right)}^{ - 2}}\left( {\dfrac{1}{x}} \right)dx = - \int {{u^{ - 2}}du} }
Now, integrate using this rule i.e.,
∫undu=un+1n+1+C\int {{u^n}du = \dfrac{{{u^{n + 1}}}}{{n + 1}} + C} , where n≠1n \ne 1.
As per the rule, we have here n = 2, in the given function. Thus, applying this to the equation, we get:
−∫u−2du=−(u−2+1−2+1)+C- \int {{u^{ - 2}}du = - \left( {\dfrac{{{u^{ - 2 + 1}}}}{{ - 2 + 1}}} \right) + C}
Simplifying the terms, we get
−∫u−2du=−(u−1−1)+C- \int {{u^{ - 2}}du = - \left( {\dfrac{{{u^{ - 1}}}}{{ - 1}}} \right) + C}
\Rightarrow $$$$ - \int {{u^{ - 2}}du = \dfrac{1}{u} + C}
Since u=ln⁡xu = \ln x,
= 1ln⁡x+C\dfrac{1}{{\ln x}} + C

Therefore, after integration we get −1x(ln⁡x)2 - \dfrac{1}{{x{{\left( {\ln x} \right)}^2}}} = 1ln⁡x+C\dfrac{1}{{\ln x}} + C

Note: To integrate the given function, we must know the derivative of the functions, such that based on its derivative we need to apply integration to the terms. There are different integration methods that are used to find an integral of some function, which is easier to evaluate the original integral. Hence, based on the function given we can find the integration of the function i.e., by using the integration methods as the details are given as additional information.