Question

How do you integrate \[{e^{ - \ln x}}\] ?

Answer 1

Hint: In order to define the integral of the above expression, we first have to rewrite the question by using the property of logarithm \[n\log m = \log {m^n}\] and consider the fact that the number $e$and $\log $are actually inverses of each other. \[{e^{\log (n)}} = n\] .By applying these properties you will get $\dfrac{1}{x}$and the integration of $\dfrac{1}{x}$ is equal to$\ln \left| x \right| + C$.

Complete step-by-step answer:
We are given an indefinite integral \[\int {{e^{ - \ln x}}dx} \]
 \[I = \int {{e^{ - \ln x}}dx} \]
To solve the given question, we must know the properties of logarithms and with the help of them we are going to rewrite our question.
But we also need to know that the number $e$and $\log $are actually inverses of each other.
First, we are going to rewrite the number with the help of the following properties of natural logarithms.
 \[
  n\log m = \log {m^n} \\
  {e^{\log (n)}} = n \;
 \]
So, \[{e^{ - \ln x}}\] can be written as
 \[{e^{ - \ln x}} = {e^{\ln {x^{ - 1}}}}\]
Let ${x^{ - 1}}$be n
Hence we can write \[{e^{\ln n}} = n = {x^{ - 1}}\]
$\therefore {e^{ - \ln x}} = {x^{ - 1}}$
Replacing this value in our original integral, we get
 \[
  I = \int {{x^{ - 1}}dx} \\
  I = \int {\dfrac{1}{x}dx} \;
 \]
And we know the rule of integration $\int {\dfrac{1}{x}dx} = \ln \left| x \right| + C$where C is the constant of integration
 \[I = \ln \left| x \right| + C\]
Therefore, the integral of \[{e^{ - \ln x}}\] is equal to \[\ln \left| x \right| + C\] .
So, the correct answer is “ \[\ln \left| x \right| + C\] .”.

Note: 1.Different types of methods of Integration:
Integration by Substitution
Integration by parts
Integration of rational algebraic function by using partial fraction
2. Integration by Substitution: The method of evaluating the integral by reducing it to standard form by a proper substitution is called integration by substitution.
If $\varphi (x)$is a continuously differentiable function, then to evaluate integrals of the form.
 \[\int {f(\varphi (x))\,{\varphi ^1}(x)dx} \] , we substitute $\varphi (x)$=t and ${\varphi ^1}(x)dx = dt$
This substitution reduces the above integral to \[\int {f(t)\,dt} \] . After evaluating this integral we substitute back the value of t.
3. Value of the constant ”e” is equal to 2.71828.
4. A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number , we actually undo an exponentiation.