Question #45e53

`int1/((x-1)^3*(x+2)^5)^(1/4)`

=`int (dx)/[(x-1)^(3/4)*(x+2)^(5/4)]`

After using `u=((x-1)/(x+2))^(1/4)` substitution,

`u^4=(x-1)/(x+2)`

`u^4*(x+2)=x-1`

`u^4*x+2u^4=x-1`

`2u^4+1=x-u^4*x`

`x=(2u^4+1)/(1-u^4)` and,

`dx=(8u^3*(1-u^4)-(-4u^3)*(2u^4+1))/(1-u^4)^2*du`

=`(12u^3*du)/(1-u^4)^2`

Also denominator became,

`(x-1)^(3/4)*(x+2)^(5/4)`

=`((2u^4+1)/(1-u^4)-1)^(3/4)*((2u^4+1)/(1-u^4)+2)^(5/4)`

=`(9u^3)/(1-u^4)^2`

Thus,

`int (dx)/[(x-1)^(3/4)*(x+2)^(5/4)]`

`int ((12u^3*du)/(1-u^4)^2)/((9u^3)/(1-u^4)^2)`

=`int 4/3*du`

=`4/3*u+C`

=`4/3*((x-1)/(x+2))^(1/4)+C`

Here is another Method to find the Integral.

Let, `I=int1/{(x-1)^3(x+2)^5}^(1/4)dx.`

`:. I=int1/[{(x-1)^4/(x-1)}{(x+2)^4(x+2)}]^(1/4)dx,`

`=int[{1/((x-1)^4(x+2)^4)^(1/4)}{(x-1)/(x+2)}^(1/4)]dx,`

`=int{1/((x-1)(x+2))*((x-1)/(x+2))^(1/4)}dx.`

Now, we substitute `(x-1)/(x+2)=t^4.`

`:.d{(x-1)/(x+2)}=d(t^4), i.e.,`

`[{(x+2)*1-(x-1)*1}/(x+2)^2]dx=4t^3dt, or,`

`3/(x+2)^2dx=4t^3dt rArrdx=4/3t^3(x+2)^2dt.`#

Also, `(x-1)/(x+2)=t^4 rArr (x-1)=t^4(x+2).`

`:. I=int{1/(t^4(x+2)^2)*(t^4)^(1/4)}4/3*t^3(x+2)^2dt,`

`=4/3int1dt,`

`=4/3t.`

`rArr I=4/3((x-1)/(x-2))^(1/4)+C, or, 4/3root(4)((x-1)/(x-2))+C.`

Enjoy Maths.!