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Question Number 32914 by mondodotto@gmail.com last updated on 06/Apr/18

Commented by mrW2 last updated on 06/Apr/18

$$\left(b\right)$$$$\int \frac{\sin x+\cos x}{\cos x-\sin x}dx$$$$=-\int \frac{\cos x+\sin x}{\sin x-\cos x}dx$$$$=-\int \frac{1}{\sin x-\cos x}d(\sin x-\cos x)$$$$=-\ln \mid \sin x-\cos x\mid +C$$

Commented by prof Abdo imad last updated on 06/Apr/18

$$a)letusethechangementx=5sht$$$$I=\int 5cht.5chtdt=25\int {ch}^{2}tdt$$$$=\frac{25}{2}\int (1+ch\left(2t\right))dt=\frac{25t}{2}+\frac{25}{2}\int ch\left(2t\right)dt$$$$=\frac{25}{2}t+\frac{25}{4}sh\left(2t\right)+\lambda butt=argsh\left(\frac{x}{5}\right)$$$$=ln(\frac{x}{5}+\sqrt{1+{\left(\frac{x}{5}\right)}^2})$$$$sh\left(2t\right)=2sh\left(t\right)ch\left(t\right)$$$$=2\frac{x}{5}\sqrt{1+{\left(\frac{x}{5}\right)}^2}\Rightarrow$$$$I=\frac{25}{2}ln(\frac{x}{5}+\sqrt{1+\frac{x^2}{25}})+\frac{5}{2}x\sqrt{1+\frac{x^2}{25}}+\lambda .$$

Commented by mondodotto@gmail.com last updated on 06/Apr/18

$$\mathbf{thanx}$$

Commented by prof Abdo imad last updated on 07/Apr/18

$$c)letputI=\int \frac{dx}{2x^2+x-3}$$$$rootsof2x^2+x-3\Rightarrow \mathrm{\Delta }=1-4\left(2\right)(-3)=25$$$$x_1=\frac{-1+5}{4}=1and{x}_2=\frac{-1-5}{2}=-3so$$$$F\left(x\right)=\frac{1}{2x^2+x-3}=\frac{1}{2(x-1)(x+3)}$$$$=\frac{a}{x-1}+\frac{b}{x+3}$$$$a={lim}_{x\to 1}(x-1)F\left(x\right)=\frac{1}{8}$$$$b={lim}_{x\to -3}(x+3)F\left(x\right)=-\frac{1}{8}\Rightarrow$$$$F\left(x\right)=\frac{1}{8(x-1)}-\frac{1}{8(x+3)}$$$$I=\frac{1}{8}\int (\frac{1}{x-1}-\frac{1}{x+3})dx+\lambda$$$$I=\frac{1}{8}ln\mid \frac{x-1}{x+3}\mid +\lambda .$$

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