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Question Number 183844 by Michaelfaraday last updated on 30/Dec/22

Answered by MJS_new last updated on 31/Dec/22

$$\int \frac{dx}{\sqrt{2x-1}+\sqrt{x-3}}=$$$$=\int \frac{\sqrt{2x-1}-\sqrt{x-3}}{(\sqrt{2x-1}+\sqrt{x-3})\left(\sqrt{2x-1-}\sqrt{x-3}\right)}dx=$$$$=\int \frac{\sqrt{2x-1}}{x+2}dx-\int \frac{\sqrt{x-3}}{x+2}dx=$$$$\mathrm{with}u=\sqrt{2x-1}\mathrm{and}v=\sqrt{x-3}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{get}$$$$=2\sqrt{2x-1}-2\sqrt{5}\arctan \frac{\sqrt{2x-1}}{\sqrt{5}}-2\sqrt{x-3}+2\sqrt{5}\arctan \frac{\sqrt{x-3}}{\sqrt{5}}+C$$

Answered by MJS_new last updated on 31/Dec/22

$$\int \frac{dx}{{4\cos }^{2}x-{5\sin }^{2}x}=\int \frac{dx}{4-{9\sin }^{2}x}=$$$$[t=\tan x\to dx=\frac{dt}{t^2+1}]$$$$=\int \frac{dt}{4-5t^2}=\frac{\sqrt{5}}{20}\ln \frac{\sqrt{5}t+2}{\sqrt{5}t-1}=$$$$=\frac{\sqrt{5}}{20}\ln \mid \frac{\sqrt{5}\sin x+2\cos x}{\sqrt{5}\sin x-2\cos x}\mid +C$$

Commented by Michaelfaraday last updated on 31/Dec/22

$$thankssir$$

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