Pleas, how solve this integral ?

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Pleas, how solve this integral ?

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Answer 1 · 2018-06-24T03:22:52

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

Explanation:

Given

$I = \int_{4}^{7} \sqrt{\frac{x + 2}{x - 3}} d x$ $I=int_4^7sqrt((x+2)/(x-3))dx$

Apply the substitution $x - 3 = u$ $x-3=u$ :

$I = \int_{1}^{4} \sqrt{\frac{u + 5}{u}} d u$ $I=int_1^4sqrt((u+5)/u)du$

Multiply numerator and denominator by $\sqrt{u + 5}$ $sqrt(u+5)$ :

$I = \int_{1}^{4} \frac{u + 5}{\sqrt{u^{2} + 5 u}} d x$ $I=int_1^4(u+5)/sqrt(u^2+5u)dx$

Rearrange:

$I = \frac{1}{2} \int_{1}^{4} \frac{2 u + 5}{\sqrt{u^{2} + 5 u}} d u + \frac{5}{2} \int_{1}^{4} \frac{1}{\sqrt{u^{2} + 5 u}} d u$ $I=1/2int_1^4(2u+5)/sqrt(u^2+5u)du+5/2int_1^4 1/sqrt(u^2+5u)du$

Complete the square in the denominator:

$I = {[\sqrt{u^{2} + 5 u}]}_{1}^{4} + 5 \int_{1}^{4} \frac{1}{\sqrt{{(2 u + 5)}^{2} - 25}} d u$ $I=[sqrt(u^2+5u)]_1^4+5int_1^4 1/sqrt((2u+5)^2-25)du$

Apply the substitution $2 u + 5 = 5 sec θ$ $2u+5=5sectheta$ :

$I = 6 - \sqrt{6} + \frac{5}{2} \int sec θ d θ$ $I=6-sqrt6+5/2intsecthetad theta$

Integrate directly:

$I = 6 - \sqrt{6} + \frac{5}{2} [ln | 5 sec θ + 5 tan θ |]$ $I=6-sqrt6+5/2[ln|5sectheta+5tantheta|]$

Reverse the last substitution:

$I = 6 - \sqrt{6} + \frac{5}{2} {[ln ∣ ∣ ∣ (2 u + 5) + \sqrt{{(2 u + 5)}^{2} - 25} ∣ ∣ ∣]}_{1}^{4}$ $I=6-sqrt6+5/2[ln|(2u+5)+sqrt((2u+5)^2-25)|]_1^4$

Insert the limits of integration:

$I = 6 - \sqrt{6} + \frac{5}{2} ln (\frac{25}{7 + 2 \sqrt{6}})$ $I=6-sqrt6+5/2ln(25/(7+2sqrt6))$

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Pleas, how solve this integral ?

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