Wednesday, March 11, 2015

int 1 / (xsqrt(x^4-4)) dx Find the indefinite integral

For the given integral: int 1/(xsqrt(x^4-4))dx , we may apply u-substitution by letting:
u =x^4-4 then  du = 4x^3 dx .
Rearrange du = 4x^3 dx into (du)/( 4x^3)= dx
Plug-in u =x^4-4  and (du)/( 4x^3)= dx , we get:
int 1/(xsqrt(x^4-4))dx =int 1/(xsqrt(u))* (du)/( 4x^3)
                        =int 1/(4x^4sqrt(u))du
Recall u =x^4-4 then adding 4 on both sides becomes: u + 4 = x^4 .
Plug-in x^4 =u+4 in the integral:
int 1/(4x^4sqrt(u))du =int 1/(4(u+4)sqrt(u))du
Apply the basic integration property: int c*f(x) dx = c int f(x) dx :
int 1/(4(u+4)sqrt(u))du=1/4int 1/((u+4)sqrt(u))du
Apply another set of substitution by letting:
v =sqrt(u)  which is the same as v^2 =u .
Then taking the derivative on both sides, we get 2v dv = du .
Plug-in u =v^2 , du = 2v dv , and sqrt(u)=v  , we get:
1/4 int 1/((u+4)sqrt(u))du = 1/4int 1/((v^2+4)v)(2v dv)
We simplify by cancelling out common factors v and 2:
1/4int 1/((v^2+4)v)(2v dv) =1/2int (dv)/(v^2+4) or1/2int (dv)/(v^2+2^2)  
                                         
The integral part resembles the integration formula:
int (du)/(u^2+a^2) = (1/a) arctan (u/a) +C
 Then, 
1/2 int (dv)/(v^2+4) =1/2 *(1/2) arctan (v/2) +C
                        =1/4 arctan (v/2) +C
Recall that we let v =sqrt(u) and u =x^4-4  then   v = sqrt(x^4-4)
Plug-in v = sqrt(x^4-4) in  1/4 arctan (v/2) +C  to get the final answer:
int 1/(xsqrt(x^4-4))dx =1/4 arctan (sqrt(x^4-4)/2) +C

No comments:

Post a Comment

Why is the fact that the Americans are helping the Russians important?

In the late author Tom Clancy’s first novel, The Hunt for Red October, the assistance rendered to the Russians by the United States is impor...