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Question Number 154880 by saly last updated on 22/Sep/21

Answered by Ar Brandon last updated on 22/Sep/21

$$I_1=\int \frac{\sqrt{x^2+9}}{x^3}dx,x=3\mathrm{sh}\vartheta$$$$=\frac{1}{3}\int \frac{{\mathrm{ch}}^2\vartheta }{{\mathrm{sh}}^3\vartheta }d\vartheta =\frac{1}{3}\int {\mathrm{cth}}^2\vartheta \cdot \mathrm{csch}\vartheta d\vartheta$$$$=-\frac{1}{3}\int \left(\sqrt{{\mathrm{csch}}^2\vartheta +1}\right)d\left(\mathrm{csch}\vartheta \right)$$$$=\frac{1}{3}\int \sqrt{u^2+1}d\left(u\right)=\frac{1}{3}\int {\mathrm{ch}}^2\theta d\theta =\frac{1}{6}(\theta +\frac{\mathrm{sh}\theta }{2})+C$$$$=\frac{1}{6}(\mathrm{argsh}\left(\mathrm{csch}\left(\mathrm{argh}\left(\frac{x}{3}\right)\right)\right)+\frac{1}{2}\mathrm{sh}\left(\mathrm{argsh}\left(\mathrm{csch}\left(\mathrm{argh}\left(\frac{x}{3}\right)\right)\right)\right)+C$$

Answered by maged last updated on 23/Sep/21

$$solution\left(1\right)$$$$I=\int \frac{xdx}{\sqrt{x^2-7}}\stackrel{x=\sqrt{7}\sec u}{=}7\int \frac{{\sec }^{2}u\tan udu}{\sqrt{7({\sec }^{2}u-1)}}$$$$=\sqrt{7}\int \frac{{\sec }^{2}u\tan udu}{\tan u}$$$$=\sqrt{7}\int {\sec }^{2}udu=\sqrt{7}\sec u\sin u+C$$$$=\sqrt{x^2-7}+C$$$$$$

Answered by Ar Brandon last updated on 23/Sep/21

$$I_1=\int \frac{\sqrt{x^2-9}}{x^3}dx,x=3\sec \vartheta \Rightarrow dx=3\sec \vartheta \tan \vartheta d\vartheta$$$$=\frac{1}{3}\int \frac{{\tan }^2\vartheta }{{\sec }^2\vartheta }d\vartheta =\frac{1}{3}\int {\sin }^2\vartheta d\vartheta =\frac{1}{6}(\vartheta -\frac{\sin 2\vartheta }{2})+\mathrm{C}$$$$=\frac{1}{6}({\sec }^{-1}\left(\frac{x}{3}\right)-\frac{1}{2}\sin \left({2\sec }^{-1}\left(\frac{x}{3}\right)\right))+C$$$$$$$$I_2=\int \frac{x}{\sqrt{x^2-7}}dx=\frac{1}{2}\int \frac{2x}{\sqrt{x^2-7}}dx$$$$=\frac{1}{2}\int \frac{d(x^2-7)}{\sqrt{x^2-7}}=\sqrt{x^2-7}+C_2$$$$$$$$I_3=\int \frac{\sqrt{x^2+4x-5}}{x+2}dx=\int \frac{\sqrt{{(x+2)}^2-9}}{x+2}dx,x+2=3\sec \vartheta$$$$=3\int {\tan }^2\vartheta d\vartheta =3\int ({\sec }^2\vartheta -1)d\vartheta =3(\tan \vartheta -\vartheta )+\mathrm{C}$$$$=3(\tan \left({\sec }^{-1}\left(\frac{x+2}{3}\right)\right)-{\sec }^{-1}\left(\frac{x+2}{3}\right))+C_3$$$$$$$$I_4=\int \frac{x^3}{{(x^2-16)}^{3/2}}dx,x=4\sec \vartheta \Rightarrow dx=4\sec \vartheta \tan \vartheta$$$$=\int \frac{{64\sec }^3\vartheta \cdot 4\sec \vartheta \tan \vartheta }{{({16\sec }^2\vartheta -16)}^{3/2}}d\vartheta =4\int \frac{{\sec }^4\vartheta \tan \vartheta }{{\tan }^3\vartheta }d\vartheta$$$$=4\int \frac{{\sec }^4\vartheta }{{\tan }^2\vartheta }d\vartheta =4\int \frac{1+{\tan }^2\vartheta }{{\tan }^2\vartheta }d\left(\tan \vartheta \right)$$$$=4(\tan \vartheta -\frac{1}{\tan \vartheta })+C=4(\tan \left({\sec }^{-1}\left(\frac{x}{4}\right)\right)+\cot \left({\sec }^{-1}\left(\frac{x}{4}\right)\right))+C_4$$

Commented by saly last updated on 24/Sep/21

$$thankyou$$

Answered by maged last updated on 23/Sep/21

$$solution:$$$$I=\int \frac{\sqrt{x^2-9}}{x^3}dx\stackrel{x=3\sec u}{=}\frac{1}{3}\int \frac{{\tan }^{2}u\sec u}{{\sec }^{3}u}du$$$$=\frac{1}{3}\int {\cos }^{2}u{\tan }^{2}udu$$$$=\frac{1}{3}\int {\sin }^{2}udu=\frac{1}{6}\int (1-\cos 2u)du$$$$=\frac{1}{6}(u-\frac{1}{2}\sin 2u)+C$$$$=\frac{1}{6}(u-\sin u\cos u)+C$$$$=\frac{1}{6}{\sec }^{-1}\left(\frac{x}{3}\right)-\frac{\sqrt{x^2-9}}{2x^2}+C$$

Answered by maged last updated on 23/Sep/21

$$I_3=\int \frac{\sqrt{x^2+4x-5}}{x+2}dx$$$$solution:$$$$I_3=\int \frac{\sqrt{x^2+4x-5}}{x+2}dx$$$$=\int \frac{\sqrt{x^2+4x+4-9}}{x+2}dx$$$$=\int \frac{\sqrt{{(x+2)}^2-9}}{x+2}dx\stackrel{x+2=3\sec u}{=}\int \frac{\sqrt{9({\sec }^{2}u-1)}}{3\sec u}3\sec u\tan udu$$$$=3\int {\tan }^{2}udu=3\int ({\sec }^{2}u-1)du$$$$=3\tan u-3u+C$$$$=\sqrt{{(x+2)}^2-9}-{3\sec }^{-1}\left(\frac{x+2}{3}\right)+C$$$$=\sqrt{x^2+4x-5}-{3\sec }^{-1}\left(\frac{x+2}{3}\right)+C$$$$$$

Commented by saly last updated on 24/Sep/21

$$thankyou$$

Answered by maged last updated on 23/Sep/21

$$I_4=\int \frac{x^3}{\sqrt{{(x^2-16)}^3}}dx$$$$solution:$$$$I=\int \frac{x^3}{\sqrt{{(x^2-16)}^3}}dx\stackrel{x=4\sec u}{=}\int \frac{{64\sec }^{3}u}{\sqrt{{16}^3{({\sec }^2-1)}^3}}4\sec u\tan udu$$$$=4^3\int \frac{{\sec }^{3}u}{4^3{\tan }^{3}u}4\sec u\tan udu$$$$=4\int \frac{{\sec }^{4}u}{{\tan }^{2}u}du=4\int \frac{{(1+{\tan }^{2}u)}^2}{{\tan }^{2}u}du$$$$=4\int {\left(\frac{1+{\tan }^{2}u}{\tan u}\right)}^{2}du=4\int {(\frac{1}{\tan u}+\tan u)}^{2}du$$$$=4\int ({\cot }^{2}u+2+{\tan }^{2}u)du$$$$=4\int ({\mathrm{cosec}}^{2}u-1+{\sec }^{2}u-1+2)du$$$$=4\int ({\sec }^{2}u+{\mathrm{cosec}}^{2}u)du$$$$=4\tan u-4\cot u+C$$$$\because x=4\sec u\to \tan u=\frac{1}{4}\sqrt{x^2-16}$$$$\cot u=\frac{4}{\sqrt{x^2-16}}$$$$I=\sqrt{x^2-16-}\frac{16}{\sqrt{x^2-16}}+C$$

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