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Question Number 208316 by Tawa11 last updated on 11/Jun/24

$$\int \frac{{\mathrm{x}}^2+3}{{\mathrm{x}}^2(\mathrm{x}+1){({\mathrm{x}}^2+1)}^2}\mathrm{dx}$$

Answered by Frix last updated on 11/Jun/24

$$\int \frac{x^2+3}{x^2(x+1){(x^2+1)}^2}dx=$$$$=-3\int \frac{dx}{x}+=-3\ln \mid x\mid -$$$$+3\int \frac{dx}{x^2}+-\frac{3}{x}+$$$$+\int \frac{dx}{x+1}++\ln \mid x+1\mid +$$$$+\int \frac{2x}{x^2+1}dx-+\ln (x^2+1)-$$$$-\frac{5}{2}\int \frac{dx}{x^2+1}+-\frac{5}{2}{\tan }^{-1}x-$$$$+\frac{1}{2}\int \frac{x^2+2x-1}{{(x^2+1)}^2}dx-\frac{x+1}{2(x^2+1)}=$$$$=-\frac{(7x^2+x+6)}{2x(x^2+1)}+\ln \mid \frac{(x+1)(x^2+1)}{x^3}\mid -\frac{5}{2}{\tan }^{-1}x+C$$

Commented by Frix last updated on 11/Jun/24

$$\frac{1}{2}\int \frac{x^2+2x-1}{{(x^2+1)}^2}dx\stackrel{t={\tan }^{-1}x}{=}$$$$=\int (\frac{1}{2}+\cos t\sin t-{\cos }^{2}t)dt=$$$$=-\frac{\sqrt{2}}{2}\int \cos (2t+\frac{\pi }{4})dt=-\frac{\sqrt{2}}{4}\sin (2t+\frac{\pi }{4})=$$$$=\frac{x^2-2x-1}{4(x^2+1)}=\frac{1}{4}-\frac{x+1}{2(x^2+1)}=-\frac{x+1}{2(x^2+1)}+C$$

Commented by Tawa11 last updated on 11/Jun/24

$$\mathrm{Thanks}\mathrm{sir}.$$$$\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{sir}.$$

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