Question #45e53

2 Answers
Nov 19, 2017

int1/((x-1)^3*(x+2)^5)^(1/4)=4/3*((x-1)/(x+2))^(1/4)+C1((x1)3(x+2)5)14=43(x1x+2)14+C

Explanation:

int1/((x-1)^3*(x+2)^5)^(1/4)1((x1)3(x+2)5)14

=int (dx)/[(x-1)^(3/4)*(x+2)^(5/4)]dx(x1)34(x+2)54

After using u=((x-1)/(x+2))^(1/4)u=(x1x+2)14 substitution,

u^4=(x-1)/(x+2)u4=x1x+2

u^4*(x+2)=x-1u4(x+2)=x1

u^4*x+2u^4=x-1u4x+2u4=x1

2u^4+1=x-u^4*x2u4+1=xu4x

x=(2u^4+1)/(1-u^4)x=2u4+11u4 and,

dx=(8u^3*(1-u^4)-(-4u^3)*(2u^4+1))/(1-u^4)^2*dudx=8u3(1u4)(4u3)(2u4+1)(1u4)2du

=(12u^3*du)/(1-u^4)^212u3du(1u4)2

Also denominator became,

(x-1)^(3/4)*(x+2)^(5/4)(x1)34(x+2)54

=((2u^4+1)/(1-u^4)-1)^(3/4)*((2u^4+1)/(1-u^4)+2)^(5/4)(2u4+11u41)34(2u4+11u4+2)54

=(9u^3)/(1-u^4)^29u3(1u4)2

Thus,

int (dx)/[(x-1)^(3/4)*(x+2)^(5/4)]dx(x1)34(x+2)54

int ((12u^3*du)/(1-u^4)^2)/((9u^3)/(1-u^4)^2)12u3du(1u4)29u3(1u4)2

=int 4/3*du43du

=4/3*u+C43u+C

=4/3*((x-1)/(x+2))^(1/4)+C43(x1x+2)14+C

Dec 5, 2017

4/3((x-1)/(x-2))^(1/4)+C, or, 4/3root(4)((x-1)/(x-2))+C.43(x1x2)14+C,or,434x1x2+C.

Explanation:

Here is another Method to find the Integral.

Let, I=int1/{(x-1)^3(x+2)^5}^(1/4)dx.I=1{(x1)3(x+2)5}14dx.

:. 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.!