Question

Pleas, how solve this integral ?

Answer 1

Use the substitutions `x-3=u` and `2u+5=5sectheta`.

Explanation:

Given

`I=int_4^7sqrt((x+2)/(x-3))dx`

Apply the substitution `x-3=u`:

`I=int_1^4sqrt((u+5)/u)du`

Multiply numerator and denominator by `sqrt(u+5)`:

`I=int_1^4(u+5)/sqrt(u^2+5u)dx`

Rearrange:

`I=1/2int_1^4(2u+5)/sqrt(u^2+5u)du+5/2int_1^4 1/sqrt(u^2+5u)du`

Complete the square in the denominator:

`I=[sqrt(u^2+5u)]_1^4+5int_1^4 1/sqrt((2u+5)^2-25)du`

Apply the substitution `2u+5=5sectheta`:

`I=6-sqrt6+5/2intsecthetad theta`

Integrate directly:

`I=6-sqrt6+5/2[ln|5sectheta+5tantheta|]`

Reverse the last substitution:

`I=6-sqrt6+5/2[ln|(2u+5)+sqrt((2u+5)^2-25)|]_1^4`

Insert the limits of integration:

`I=6-sqrt6+5/2ln(25/(7+2sqrt6))`