Matematické Fórum

stuart clark · 25. 04. 2018 02:14

$\int\frac{(x-2)}{(7x^2-36x+48)\sqrt{x^2-2x-1}}dx$

Options:

$(a)\;\; -2\tan^{-1}\bigg(\frac{\sqrt{3x^2-6x-12}}{9-3x}\bigg)+\mathcal{C}$

$(b)\;\; -2\tan^{-1}\bigg(\frac{\sqrt{4x^2-6x-12}}{9-3x}\bigg)+\mathcal{C}$

$(c)\;\; -2\tan^{-1}\bigg(\frac{\sqrt{3x^2-6x-10}}{9-3x}\bigg)+\mathcal{C}$

$(d)\;\; -2\tan^{-1}\bigg(\frac{\sqrt{3x^2-6x-1}}{9-3x}\bigg)+\mathcal{C}$

stuart clark

Thanks friends got it

put $x=\frac{x-2}{3-x}$ , Then $x=\frac{2+3t}{t+1}=1-\frac{1}{(t+1)}$

and $dx=\frac{1}{(t+1)^2}$ and $(x-2)=\frac{t}{t+1}$ and $x^2-2x-1=\frac{2t^2-1}{(t+1)^2}$

and $7x^2-36x+48=\frac{3t^2+4}{(t+1)^2}$

so $I=\int\frac{t}{(3t^2+4)\sqrt{2t^2-1}}dt$

Substitute $(2t^2-1)=u^2$ and $2tdt=udu$

So $\frac{1}{2}\int\frac{1}{3u^2+11}du=\frac{1}{\sqrt{33}}\tan^{-1}\bigg(\frac{u\sqrt{3}}{\sqrt{11}}\bigg)+\mathcal{C}$

Matematické Fórum

#1 25. 04. 2018 02:14

Irrational Indefinite Integral

#2 17. 05. 2018 08:45 — Editoval stuart clark (17. 05. 2018 08:45)

Re: Irrational Indefinite Integral

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