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Question Number 145671 by mathlove last updated on 07/Jul/21

Commented by mathlove last updated on 07/Jul/21

$$pleashelpme$$

Answered by imjagoll last updated on 07/Jul/21

$$\left(11\right)\int \frac{\sqrt{{\mathrm{x}}^2-9}}{({\mathrm{x}}^2+3\mathrm{x})(\mathrm{x}-3)}\mathrm{dx}$$$$=\int \frac{\sqrt{\mathrm{x}+3}\sqrt{\mathrm{x}-3}}{\mathrm{x}{\left(\sqrt{\mathrm{x}+3}\right)}^2{\left(\sqrt{\mathrm{x}-3}\right)}^2}\mathrm{dx}$$$$=\int \frac{\mathrm{dx}}{\mathrm{x}\sqrt{{\mathrm{x}}^2-9}}$$$$\mathrm{let}\mathrm{x}=3\sec \mathrm{t}\Rightarrow \mathrm{dx}=3\sec \mathrm{t}\tan \mathrm{t}\mathrm{dt}$$$$\mathrm{I}=\int \frac{3\sec \mathrm{t}\tan \mathrm{t}\mathrm{dt}}{3\sec \mathrm{t}\sqrt{9\tan {}^2\mathrm{t}}}$$$$\mathrm{I}=\int \frac{1}{3}\mathrm{dt}=\frac{1}{3}\mathrm{arcsec}\left(\frac{\mathrm{x}}{3}\right)+\mathrm{c}$$

Commented by mathlove last updated on 07/Jul/21

$$andohter$$

Answered by puissant last updated on 07/Jul/21

$$16)$$$$\mathrm{t}={\mathrm{x}}^2\Rightarrow \mathrm{dt}=2\mathrm{xdx}\Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}}$$$$\mathrm{I}=\int \frac{{\mathrm{x}}^3}{{({\mathrm{x}}^2+1)}^3}\times \frac{\mathrm{dt}}{2\mathrm{x}}=\frac{1}{2}\int \frac{{\mathrm{x}}^2}{{({\mathrm{x}}^2+1)}^3}\mathrm{dt}$$$$=\frac{1}{2}\int \frac{\mathrm{t}}{{(\mathrm{t}+1)}^3}\mathrm{dt}=\frac{1}{2}\int \frac{1}{{(\mathrm{t}+1)}^2}\mathrm{dt}-\frac{1}{2}\int \frac{1}{{(\mathrm{t}+1)}^3}\mathrm{dt}$$$$\Rightarrow \mathrm{I}=-\frac{1}{2(\mathrm{t}+1)}+\frac{1}{4{(\mathrm{t}+1)}^2}+\mathrm{C}$$$$\mathrm{I}=\frac{1}{4{({\mathrm{x}}^2+1)}^2}-\frac{1}{2({\mathrm{x}}^2+1)}+\mathrm{C}..$$

Commented by mathlove last updated on 07/Jul/21

$$andother$$

Answered by iloveisrael last updated on 07/Jul/21

$$\left(9\right)\int \frac{\cos 3u}{{(4+4\sin 3u+\sin {}^{2}3u)}^2}du$$$$=\int \frac{\cos 3u}{{(2+\sin 3u)}^2}du$$$$=\frac{1}{3}\int \frac{d(2+\sin 3u)}{{(2+\sin 3u)}^2}$$$$=-\frac{1}{3(2+\sin 3u)}+c$$

Commented by mathlove last updated on 08/Jul/21

$$thankssirandother$$

Answered by bobhans last updated on 07/Jul/21

$$\left(5\right)\int \frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{2\mathrm{x}}}{\sqrt{{\mathrm{e}}^{\mathrm{x}}+1}}\mathrm{dx}=\int \frac{{\mathrm{e}}^{\mathrm{x}}({\mathrm{e}}^{\mathrm{x}}+1)}{\sqrt{{\mathrm{e}}^{\mathrm{x}}+1}}\mathrm{dx}$$$$=\int {\mathrm{e}}^{\mathrm{x}}\sqrt{1+{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx}=\int \sqrt{1+{\mathrm{e}}^{\mathrm{x}}}\mathrm{d}(1+{\mathrm{e}}^{\mathrm{x}})$$$$=\frac{2}{3}\sqrt{{(1+{\mathrm{e}}^{\mathrm{x}})}^3}+\mathrm{c}$$

Answered by bobhans last updated on 07/Jul/21

$$\left(15\right)\int \frac{{(3+\mathrm{x})}^2-6\mathrm{x}}{\sqrt{{(9+{\mathrm{x}}^2)}^5}}\mathrm{dx}=\int \frac{{\mathrm{x}}^2+9}{\sqrt{{({\mathrm{x}}^2+9)}^5}}\mathrm{dx}$$$$=\int \frac{\mathrm{dx}}{\sqrt{{({\mathrm{x}}^2+9)}^3}};\mathrm{x}=3\tan \mathrm{u};\sin \mathrm{u}=\frac{\mathrm{x}}{\sqrt{9+{\mathrm{x}}^2}}$$$$=\int \frac{3\sec {}^2\mathrm{u}}{\sqrt{{\left(9\sec {}^2\mathrm{u}\right)}^3}}=\frac{1}{9}\int \frac{\mathrm{du}}{\sec \mathrm{u}}$$$$=\frac{1}{9}\int \cos \mathrm{u}\mathrm{du}=\frac{1}{9}\sin \mathrm{u}+\mathrm{c}=\frac{\mathrm{x}}{9\sqrt{9+{\mathrm{x}}^2}}+\mathrm{c}$$

Answered by puissant last updated on 07/Jul/21

$$8)\int \frac{\left({\mathrm{e}}^2\cos \left(\mathrm{x}\right)\right)\left({\mathrm{e}}^{\sin \left(\mathrm{x}\right)}\right)}{{\mathrm{e}}^{\mathrm{lne}}+\ln 1}\mathrm{dx}$$$$=\int \frac{\cos \left(\mathrm{x}\right){\mathrm{e}}^{2+\sin \left(\mathrm{x}\right)}}{\mathrm{e}}\mathrm{dx}$$$$=\frac{1}{\mathrm{e}}\int \cos \left(\mathrm{x}\right){\mathrm{e}}^{2+\sin \left(\mathrm{x}\right)}\mathrm{dx}$$$$=\frac{1}{\mathrm{e}}{\mathrm{e}}^{2+\sin \left(\mathrm{x}\right)}+\mathrm{c}...$$

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