Anti Derivative of Log X

Introduction to anti derivative of log x:

A function `Phi (x) ` is called a primitive or an antiderivative of a function f(x) if `Phi '(x)`  = f(x).

For example , `(x^4)/4` is an antiderivative of x 3, because `d/dx (x^4)/4`  = x 3.

To find the antiderivative of log x we use a special form of antiderivatives known as uv form (antiderivative by parts).

Theorem : If u and v are two fucntions of x, then

`int` u v dx  = u `int` vdx - `int { du/dx int v dx } `  dx

Finding Antiderivative of Logx

We use the above said formula of uv form to find the antiderivative of logx.

We need two functions for that logx can be written as 1 * logx.

We take logx = u , as differentiation of log x is easy to find and v = 1.

`(du)/dx = (d(logx))/dx`  = `1/x`

and `int` v dx =  `int` 1 dx = x

We use all these values and plug it in the formula

The uv form is `int` uv dx = u `int` v dx - `int { (du)/dx int v dx }` dx

`int` 1 logx dx = logx `int` 1 dx  -  `int { (du)/dx int 1 dx }` dx

= logx * x -  `int 1/x xx x dx`  dx

= x logx - `int 1 dx`

= xlogx - x  +c

= x(logx -1) +c

The antiderivative of logx is x(logx-1) +c

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Solved Problem on Antiderivative of Logx

Find the antiderivative of xlogx   

Solution :  The function is x logx

Let v =x and u = logx, as log x is easily differentiable and the integral of x is easy to find.

Now `int` v dx  = `int` x dx = `(x^2)/2`

and `(d)/dx = (d(logx))/dx = 1/x`

Plugging in all the values in the formula

`int` uv dx  =  u `int` vdx - `int { (du)/dx int vdx }` dx

`int` x logx dx = logx `int` x dx - `int {(d(logx ))/dx int x dx}`  dx

= logx `x^2/2`  - `int` `1/x xx x^2/2` dx

= `x^2/2`  logx - `1/2 int x` dx

= `x^2/2` logx - `1/2 * x^2/2` +c

= `x^2/2`    ( logx - `1/2` ) +c

Antiderivative of xlogx is  `x^2/2`    ( logx - `1/2` ) +c