Pleas, how solve this integral ?

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Answer 1 · 2018-06-24T03:22:52

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

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:

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