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How do you integrate $\dfrac{1}{{x\ln x}}dx$ ?

seo-qna
Last updated date: 13th Jun 2024
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Answer
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Hint: For finding a very small part of a whole quantity, we use derivatives, while integration means finding the whole quantity from the given small part, integration is also called antiderivative. In this question, we have to find the integration of the given quantity. The function which has to be integrated is a fraction containing the product of x and the natural logarithm of x, so we can simplify it by using the substitution method and then find its integration.

Complete step-by-step solution:
In this question, we have to integrate $\dfrac{1}{{x\ln x}}dx$
For that we will let $\ln x = u$
Now, differentiating the above-supposed condition with respect to x, we get –
$
  \dfrac{{d(\ln x)}}{{dx}} = \dfrac{{du}}{{dx}} \\
   \Rightarrow du = \dfrac{1}{x}dx \\
 $
Put the above two values in the given equation, we get –
$
  \dfrac{1}{{x\ln x}}dx = \dfrac{1}{u}du \\
   \Rightarrow \int {\dfrac{1}{{x\ln x}}dx} = \int {\dfrac{1}{u}du} \\
   \Rightarrow \int {\dfrac{1}{{x\ln x}}dx} = \ln \left| u \right| + c \\
   \Rightarrow \int {\dfrac{1}{{x\ln x}}dx} = \ln (\left| {\ln x} \right|) + c \\
 $
Hence the integration of $\dfrac{1}{{x\ln x}}dx$ is $\ln (\left| {\ln x} \right|) + c$ .

Additional information:
In differential calculus, we have to find the derivative or differential of a given function but integration is the inverse process of differentiation. When the derivative of a function is given and we have to find the function, we use integration.

Note: There are two types of integrals, definite integral and indefinite integrals, a definite integral is defined as an integral that is expressed with upper and lower limits while an integral that is expressed without limits like in this question is known as an indefinite integral. The derivative of a function is unique but a function can have infinite integrals or anti-derivatives. Here, one can get different values of integral of a function by varying the value of the arbitrary constant.