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.!