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Question Number 128529 by bramlexs22 last updated on 08/Jan/21

Commented by liberty last updated on 08/Jan/21

$$\mathrm{considering}\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{{\mathrm{u}}^{\prime }\mathrm{v}-{\mathrm{uv}}^{\prime }}{{\mathrm{v}}^2}$$$$\mathrm{obvious}\mathrm{v}=2+3\cos \mathrm{x}\mathrm{and}3+2\cos \mathrm{x}={\mathrm{u}}^{\prime }(2+3\cos \mathrm{x})-\mathrm{u}(-3\sin \mathrm{x})$$$$\Rightarrow 2\cos \mathrm{x}+3={2\mathrm{u}}^{\prime }+{3\mathrm{u}}^{\prime }\cos \mathrm{x}+3\mathrm{u}\sin \mathrm{x}$$$$\mathrm{check}\mathrm{u}=\sin \mathrm{x}\Rightarrow {\mathrm{u}}^{\prime }(2+3\mathrm{cosx})+3\mathrm{u}\sin \mathrm{x}=$$$$2\cos \mathrm{x}+{3\cos }^2\mathrm{x}+{3\sin }^2\mathrm{x}=2\cos \mathrm{x}+3\left(\mathrm{valid}\right)$$$$\mathrm{therefore}\int \frac{3+2\cos \mathrm{x}}{{(2+3\cos \mathrm{x})}^2}\mathrm{dx}=\frac{\sin \mathrm{x}}{2+3\cos \mathrm{x}}$$$$$$

Commented by SLVR last updated on 08/Jan/21

$$sir..justmultiplyanddividewith{Cosec}^{2}x$$$$willgivethesameanswer.$$

Answered by SLVR last updated on 08/Jan/21

Commented by bramlexs22 last updated on 08/Jan/21

$$\mathrm{what}?$$

Commented by SLVR last updated on 08/Jan/21

$$ihopeareagreedwiththesolution$$

Commented by john_santu last updated on 08/Jan/21

$$\mathrm{what}\mathrm{your}\mathrm{solution}?$$

Commented by SLVR last updated on 09/Jan/21

Commented by SLVR last updated on 09/Jan/21

$$sorry...ididnotobserveduploadingpart$$

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