00:01
In this question, we are required to find the value of integration log x upon x square d x with the help of any simplest method.
00:14
So let's see how to solve this question.
00:16
First of all, let's simplify this expression.
00:19
So we can write integration x to the power minus 2, log x d x.
00:29
Now, to solve this, consider u is equal to log x and dv is equals to x to x to the power minus 2 d x.
00:44
Therefore by the differentiation of u we can write, d u is equal 2 and the differentiation of log x is 1 upon x d x.
00:54
And by the integration of dv we can write v is equal to x to the power minus 2 plus 1 divided by minus 2 plus 1.
01:06
It means v is equal to x to the power minus 1 upon minus 1 is equal to minus x to the power minus 1.
01:21
Now we have the values of u v, du and dv therefore we can apply the formula of integration by parts, which can be written as integration udv is equal to uv minus integration vdu.
01:37
Now substitute all the values...