Question

How do you find the integral `int (sqrtx-1)^2/sqrtxdx` using substitution?

Answer 1

`int (sqrtx-1)^2/sqrtxdx = 2/3(sqrt x-1)^3+C`

Explanation:

Calling `y = sqrt(x)` after

`dy =1/2 dx/sqrt(x)`

and substituting

`int (sqrtx-1)^2/sqrtxdx equiv 2int (y-1)^2 dy = 2/3(y-1)^3+C`

Recovering the initial variable

`int (sqrtx-1)^2/sqrtxdx = 2/3(sqrt x-1)^3+C`

Answer 2

`(2sqrtx)/3(x-3sqrtx+3)+C`.

Explanation:

Let `I=int(sqrtx-1)^2/sqrtxdx`

We subst. `x=t^2 rArr dx=2tdt`. Also,

`(sqrtx-1)^2/sqrtx=(t-1)^2/t=(t^2-2t+1)/t`.

`rArr I=int{(t^2-2t+1)/t}2tdt`

`=2int(t^2-2t+1)dt`

`=2[t^3/3-2t^2/2+t]`

`=(2t)/3(t^2-3t+3)`

`:. I = (2sqrtx)/3(x-3sqrtx+3)+C`.

Though we have solved the Problem using Substn. Methodas was

so demanded , but, in fact, there is no such need to solve it

in this fashion. Have a look :

`I=int(sqrtx-1)^2/sqrtxdx`

`=int{(x-2sqrtx+1)}/sqrtxdx`

`=int{x/sqrtx-2sqrtx/sqrtx+1/sqrtx}dx`

`=int(x^(1/2)-2+x^(-1/2))dx`

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

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

`=(2x^(1/2))/3(x-3x^(1/2)+3), or, (2sqrtx)/3(x-3sqrtx+3)+C`, as before!

Enjoy Maths.!