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Answer 1 · 2017-04-03T21:57:25

$\frac{1}{2} ln ∣ ∣ ∣ \frac{x - 3}{2 + \sqrt{6 x - x^{2} - 5}} ∣ ∣ ∣ + C$ $1/2lnabs((x-3)/(2+sqrt(6x-x^2-5)))+C$

$I = \int \frac{d x}{(x - 3) \sqrt{6 x - x^{2} - 5}}$ $I=intdx/((x-3)sqrt(6x-x^2-5))$

Complete the square in the square root:

$I = \int \frac{d x}{(x - 3) \sqrt{- (x^{2} - 6 x + 9) - 5 + 9}}$ $I=intdx/((x-3)sqrt(-(x^2-6x+9)-5+9))$

$I = \int \frac{d x}{(x - 3) \sqrt{4 - {(x - 3)}^{2}}}$ $I=intdx/((x-3)sqrt(4-(x-3)^2))$

Let $x - 3 = t$ $x-3=t$ . This implies that $d x = d t$ $dx=dt$ :

$I = \int \frac{d t}{t \sqrt{4 - t^{2}}}$ $I=intdt/(tsqrt(4-t^2))$

Now let $t = 2 sin θ$ $t=2sintheta$ . This implies that $d t = 2 cos θ d θ$ $dt=2costhetad theta$ .

$I = \int \frac{2 cos θ d θ}{2 sin θ \sqrt{4 - 4 {sin}^{2} θ}}$ $I=int(2costhetad theta)/(2sinthetasqrt(4-4sin^2theta))$

Note that $\sqrt{4 - 4 {sin}^{2} θ} = 2 \sqrt{1 - {sin}^{2} θ} = 2 cos θ$ $sqrt(4-4sin^2theta)=2sqrt(1-sin^2theta)=2costheta$ :

$I = \int \frac{2 cos θ d θ}{2 sin θ (2 cos θ)}$ $I=int(2costhetad theta)/(2sintheta(2costheta))$

$I = \frac{1}{2} \int csc θ d θ$ $I=1/2intcscthetad theta$

This is a well known integral:

$I = - \frac{1}{2} ln | csc θ + cot θ |$ $I=-1/2lnabs(csctheta+cottheta)$

$I = - \frac{1}{2} ln ∣ ∣ ∣ \frac{1 + cos θ}{sin θ} ∣ ∣ ∣$ $I=-1/2lnabs((1+costheta)/sintheta)$

$I = \frac{1}{2} ln ∣ ∣ ∣ ∣ \frac{sin θ}{1 + \sqrt{1 - {sin}^{2} θ}} ∣ ∣ ∣ ∣$ $I=1/2lnabs(sintheta/(1+sqrt(1-sin^2theta)))$

Our substitution $t = 2 sin θ$ $t=2sintheta$ implies that $sin θ = t / 2$ $sintheta=t//2$ :

$I = \frac{1}{2} ln ∣ ∣ ∣ ∣ ∣ \frac{t / 2}{1 + \sqrt{1 - {(t / 2)}^{2}}} ∣ ∣ ∣ ∣ ∣$ $I=1/2lnabs((t//2)/(1+sqrt(1-(t//2)^2)))$

Note that $\sqrt{1 - {(t / 2)}^{2}} = \sqrt{\frac{4 - t^{2}}{4}} = \frac{1}{2} \sqrt{4 - t^{2}}$ $sqrt(1-(t//2)^2)=sqrt((4-t^2)/4)=1/2sqrt(4-t^2)$ :

$I = \frac{1}{2} ln ∣ ∣ ∣ ∣ \frac{t / 2}{1 + \frac{1}{2} \sqrt{4 - t^{2}}} ∣ ∣ ∣ ∣$ $I=1/2lnabs((t//2)/(1+1/2sqrt(4-t^2)))$

$I = \frac{1}{2} ln ∣ ∣ ∣ \frac{t}{2 + \sqrt{4 - t^{2}}} ∣ ∣ ∣$ $I=1/2lnabs(t/(2+sqrt(4-t^2)))$

Using $t = x - 3$ $t=x-3$ :

$I = \frac{1}{2} ln ∣ ∣ ∣ \frac{x - 3}{2 + \sqrt{6 x - x^{2} - 5}} ∣ ∣ ∣ + C$ $I=1/2lnabs((x-3)/(2+sqrt(6x-x^2-5)))+C$