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...
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`
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