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Question

Question: How do you find the primitive of \({{e}^{2\log x}}\) ?...

How do you find the primitive of e2logx{{e}^{2\log x}} ?

Explanation

Solution

Here we have to find the primitive of the given function which means we have to integrate the given function. Firstly we will write the function inside an integral sign then by using exponent and logarithm relation we will simplify our value. Finally by using the basic formula for integrating a variable we will get our desired answer.

Complete step by step answer:
We have to find the primitive of the following function;
e2logx{{e}^{2\log x}}
So we have to find the ant-derivative of above value which can be written as follows:
e2logxdx\int{{{e}^{2\log x}}dx}
Now as we know the property of logarithm that alogb=logbaa\log b=\log {{b}^{a}} using it in the power of above value we get,
e2logxdx=elogx2dx\Rightarrow \int{{{e}^{2\log x}}dx}=\int{{{e}^{\log {{x}^{2}}}}dx}
Next we know the property of exponent that eloga=a{{e}^{\log a}}=a using it above we get,
e2logxdx=x2dx\Rightarrow \int{{{e}^{2\log x}}dx}=\int{{{x}^{2}}dx}….(1)\left( 1 \right)
The basic formula for finding integration of a single variable without any other function in it is given as follows:
xndx=xn+1n+1+C\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C
Where CC is any constant
Using the above formula in equation (1) where n=2n=2 we get,
e2logxdx=x2+12+1+C\Rightarrow \int{{{e}^{2\log x}}dx}=\dfrac{{{x}^{2+1}}}{2+1}+C
e2logxdx=x33+C\Rightarrow \int{{{e}^{2\log x}}dx}=\dfrac{{{x}^{3}}}{3}+C
Hence primitive of e2logx{{e}^{2\log x}} is x33+C\dfrac{{{x}^{3}}}{3}+C where CC is any constant.

Note:
Integrating any function has different names in mathematics such as antiderivative, primitive function or integral. Don’t get confused when such terms are used; they all mean that we have to find the integration of the given function. Using exponential and logarithmic relation and properties is very important in such a question as it simplifies our value which is to be integrated. Otherwise for integrating an exponential function we have to use the By-parts method which is little long and requires more calculation.