∫((xdx)/(sin^2 x−3))=? help me


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Question Number 97400 by student work last updated on 07/Jun/20

$\int \frac{xdx}{\sin^{2} x - 3} = ?$ $help me$
Commented by Tony Lin last updated on 08/Jun/20
$sinx=((e^(ix) −e^(−ix) )/(2i)) ∫(x/(−(((e^(ix) −e^(−ix) )/2))^2 −3))dx =−4∫(x/((e^(ix) −e^(−ix) )^2 +12))dx =−4∫((xe^(2ix) )/((e^(2ix) −1)^2 +12e^(2ix) ))dx let u=e^(2ix) −1, dx=(du/(2ie^(2ix) )) x=−((iln(u+1))/2) −∫((ln(u+1))/(u^2 +12+12))du =−((ln(u+1))/([u−(2(√6)−6)][u−(−2(√6)−6]))du =(1/(4(√6)))[∫((ln(u+1))/(u+2(√6)+6))du−∫((ln(u+1))/(u−2(√6)+6))du] let v=u+2(√6)+6,dv=du let t=u−2(√6)+6,dt=du (1/(4(√6)))[∫((ln(v−2(√6)−5))/v)dv−∫((ln(t+2(√6)−5))/t)dt] =(1/(4(√6)))∫[(( ln((v/(−2(√6)−5))+1))/v)+ln(−2(√6)−5)(1/v)]dv −(1/(4(√6)))∫[((ln((t/(2(√6)−5))+1))/t)+ln(2(√6)−5)(1/v)]dv let w=−(v/(−2(√6)−5)) ,dv=(5+2(√6))dw let z=−(t/(−2(√6)+5)) ,dt=(5−2(√6))dz (1/(4(√6))){−∫−((ln(1−w))/w)dw+∫−((ln(1−z))/z)dz +ln(−2(√6)−5)ln∣v∣+ln(2(√6)−5)ln∣t∣} =(1/(4(√6)))[Li_2 (w)−Li_2 (z)+ln(−2(√6)−5)ln∣u+2(√6)+6∣ +ln(2(√6)−5)ln∣u−2(√6)+6∣+c =(1/(4(√6)))[Li_2 (((u+2(√6)+6)/(2(√6)+5)))−Li_2 (((u−2(√6)+6)/(2(√6)−5))) +ln(−2(√6)−5)ln∣u+2(√6)+6∣ +ln(2(√6)−5)ln∣u−2(√6)+6∣+c =(1/(4(√6)))[Li_2 (((e^(2ix) +2(√6)+5)/(2(√6)+5)))−Li_2 (((e^(2ix) −2(√6)+5)/(2(√6)−5))) +ln(−2(√6)−5)ln∣e^(2ix) +2(√6)+5∣ +ln(2(√6)−5)ln∣e^(2ix) −2(√6)+5∣+c$
$s i n x = \frac{e^{i x} - e^{- i x}}{2 i}$ $\int \frac{x}{- {(\frac{e^{i x} - e^{- i x}}{2})}^{2} - 3} d x$ $= - 4 \int \frac{x}{{(e^{i x} - e^{- i x})}^{2} + 12} d x$ $= - 4 \int \frac{{x e}^{2 i x}}{{(e^{2 i x} - 1)}^{2} + 12 e^{2 i x}} d x$ $l e t u = e^{2 i x} - 1, d x = \frac{d u}{2 {i e}^{2 i x}}$ $x = - \frac{i l n (u + 1)}{2}$ $- \int \frac{l n (u + 1)}{u^{2} + 12 + 12} d u$ $= - \frac{l n (u + 1)}{[u - (2 \sqrt{6} - 6)] [u - (- 2 \sqrt{6} - 6]} d u$ $= \frac{1}{4 \sqrt{6}} [\int \frac{l n (u + 1)}{u + 2 \sqrt{6} + 6} d u - \int \frac{l n (u + 1)}{u - 2 \sqrt{6} + 6} d u]$ $l e t v = u + 2 \sqrt{6} + 6, d v = d u$ $l e t t = u - 2 \sqrt{6} + 6, d t = d u$ $\frac{1}{4 \sqrt{6}} [\int \frac{l n (v - 2 \sqrt{6} - 5)}{v} d v - \int \frac{l n (t + 2 \sqrt{6} - 5)}{t} d t]$ $= \frac{1}{4 \sqrt{6}} \int [\frac{l n (\frac{v}{- 2 \sqrt{6} - 5} + 1)}{v} + l n (- 2 \sqrt{6} - 5) \frac{1}{v}] d v$ $- \frac{1}{4 \sqrt{6}} \int [\frac{l n (\frac{t}{2 \sqrt{6} - 5} + 1)}{t} + l n (2 \sqrt{6} - 5) \frac{1}{v}] d v$ $l e t w = - \frac{v}{- 2 \sqrt{6} - 5}, d v = (5 + 2 \sqrt{6}) d w$ $l e t z = - \frac{t}{- 2 \sqrt{6} + 5}, d t = (5 - 2 \sqrt{6}) d z$ $\frac{1}{4 \sqrt{6}} {- \int - \frac{l n (1 - w)}{w} d w + \int - \frac{l n (1 - z)}{z} d z$ $+ l n (- 2 \sqrt{6} - 5) l n ∣ v ∣ + l n (2 \sqrt{6} - 5) l n ∣ t ∣}$ $= \frac{1}{4 \sqrt{6}} [{L i}_{2} (w) - {L i}_{2} (z) + l n (- 2 \sqrt{6} - 5) l n ∣ u + 2 \sqrt{6} + 6 ∣$ $+ l n (2 \sqrt{6} - 5) l n ∣ u - 2 \sqrt{6} + 6 ∣ + c$ $= \frac{1}{4 \sqrt{6}} [{L i}_{2} (\frac{u + 2 \sqrt{6} + 6}{2 \sqrt{6} + 5}) - {L i}_{2} (\frac{u - 2 \sqrt{6} + 6}{2 \sqrt{6} - 5})$ $+ l n (- 2 \sqrt{6} - 5) l n ∣ u + 2 \sqrt{6} + 6 ∣$ $+ l n (2 \sqrt{6} - 5) l n ∣ u - 2 \sqrt{6} + 6 ∣ + c$ $= \frac{1}{4 \sqrt{6}} [{L i}_{2} (\frac{e^{2 i x} + 2 \sqrt{6} + 5}{2 \sqrt{6} + 5}) - {L i}_{2} (\frac{e^{2 i x} - 2 \sqrt{6} + 5}{2 \sqrt{6} - 5})$ $+ l n (- 2 \sqrt{6} - 5) l n ∣ e^{2 i x} + 2 \sqrt{6} + 5 ∣$ $+ l n (2 \sqrt{6} - 5) l n ∣ e^{2 i x} - 2 \sqrt{6} + 5 ∣ + c$
Commented by student work last updated on 08/Jun/20

$thanks lot very good$
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