Integration by Parts Rule

 

Integration process is the inverse process of the differentiation process. If `dy/dx` =g(x) here y is the function then `int` g(x) dx = y. There are two types of integration namely indefinite integral (without  limits) and definite integral (with limits).

 

 

If we consider u = f(x), v = g(x), then the product rule in its simplest form is:
`int u dv/dx dx = uv - int v du/dx dx`
(Source from Wikipedia)Examples to Explain "integration by Parts Rule "
Rules for choosing the terms u and dv:
If the given function is combined with x and the exponential, then u = x and dv = exponential
If the given function is combined with x and the logarithmic, then u is logarithmic and dv = x
For example consider  `int 5x^3 ln x dx`
Solution:
Here logarithmic exists.
Take ln x as u and `x^3 dx` as dv
u =5 `ln x`                   dv =`x^3 dx`
`(du)/dx = 5/x`                   v = `int x^3 dx`
du = `5dx/x`                  v = `x^4/4 `
We have  `int u dv/dx dx = uv - int v du/dx dx`
0r `int u dv = uv - int v du`
`int 5x^3 ln x dx`  = 5 ln x . `x^4/4 - ` `int x^4/4 5dx/x`
=  5 ln x . `x^4/4 - ` `5/4int x^3 dx`
=  5 ln x . `x^4/4 - `   `5/4 (1/4 x^4) + c`
= `5 ln x (x^4)/4``-(5x^4)/16 + c`
This is the result that required after the process integration by part.
consider  `int 15xe^(13x) dx`
Solution:
Here exponential exists.
Take x as u and `e^(13x)dx` as dv
u = 15x                    dv =  `e^(13x)dx`
`du/dx = 15`             v = `int e^(13x) dx`
du = 15dx                  v = `1/13 e^(13x)`
We have  `int u dv/dx dx = uv - int v du/dx dx`
0r `int u dv = uv - int v du`
`int 15xe^(13x) dx` = 15x . `1/13 e^(13x) -`    `int 1/13 e^(13x) 15dx`
=  `(15x)/13 e^(13x) -` `15/13 int e^(13x) dx`
= `(15x)/13 e^(13x) -`  `15/13 (1/13 e^(13x)) + c`
= `(15x e^(13x))/13``-(15e^(13x))/169 + c`
This is the result that required after the process integration by part on the exponential.Problems to Explain "integration by Parts Rule "
Consider `int x log x` for integration by parts
Solution:
`int x log x` = `int ( log x) (x dx)`
log x is not to be integrable so we take it as
u = logx       and        dv = x dx
∴ du = 1/x dx              v = `x^2/2`
∴ `int x log x` = (logx) `x^2/2`  -  `int (x^2/2) (1/x dx)`
= `x^2/2`  (logx) `-1/2` `int x dx`
∴ `int x log x` =  `x^2/2`  (logx) - `1/4 x^2 + c`
This is the result that required after the process integration by part.

 

Consider `int x sec^2 x dx` for integration by parts
Solution:
`int x sec^2 x dx`   =  `int (x) (sec^2 x dx)`
Applying integration by parts, we get  dv = sec^2x dx    then  v = tanx        and  u = x     then du = dx
`int x sec^2 x dx`   =  x tanx − `int ` tanx dx
=  x tanx − log sec x + c
= x tanx + log cosx + c
This is the result that required after the process integration by part.