Question and Answers Forum

All Questions      Topic List

Vector Questions

Previous in All Question      Next in All Question      

Previous in Vector      Next in Vector      

Question Number 152588 by rexford last updated on 30/Aug/21

$${\int }_{-1}^1\frac{3x+4}{3+4x+3x^2}dt$$$$please,helpme$$

Answered by qaz last updated on 30/Aug/21

$${\int }_{-1}^{+1}\frac{3\mathrm{x}+4}{3+4\mathrm{x}+{3\mathrm{x}}^2}\mathrm{dt}=\frac{3\mathrm{x}+4}{3+4\mathrm{x}+{3\mathrm{x}}^2}\cdot \mathrm{t}{\mid }_{-1}^1=\frac{6\mathrm{x}+8}{3+4\mathrm{x}+{3\mathrm{x}}^2}$$

Answered by puissant last updated on 30/Aug/21

$$=\frac{1}{2}{\int }_{-1}^1\frac{6x+4+4}{3x^2+4x+3}dx$$$$=\frac{1}{2}{\int }_{-1}^1\frac{6x+4}{3x^2+4x+3}dx+\frac{2}{3}{\int }_{-1}^{+1}\frac{1}{x^2+\frac{4}{3}x+1}dx$$$$=\frac{1}{2}{[ln\mid 3x^2+4x+3\mid ]}_{-1}^1+\frac{2}{3}Q$$$$=\frac{1}{2}ln5+\frac{2}{3}Q$$$$Q={\int }_{-1}^1\frac{1}{x^2+\frac{4}{3}x+1}dx={\int }_{-1}^1\frac{1}{{(x+\frac{2}{3})}^2-\frac{4}{9}+\frac{9}{9}}dx$$$$={\int }_{-1}^1\frac{1}{{(x+\frac{2}{3})}^2+\frac{5}{9}}dx=\frac{9}{5}{\int }_{-1}^1\frac{1}{[{\left(\frac{3}{\sqrt{5}}(x+\frac{2}{3})\right)}^2+1]}dx$$$$u=\frac{3}{\sqrt{5}}(x+\frac{2}{3})\to dx=\frac{\sqrt{5}}{3}du$$$$\Rightarrow Q=\frac{9}{5}\times \frac{\sqrt{5}}{3}{\int }_{-1}^1\frac{1}{u^2+1}du$$$$=\frac{3}{\sqrt{5}}{\left[arctan\left(u\right)\right]}_{-1}^1=\frac{3}{\sqrt{5}}\times \frac{\pi }{2}.$$$$\therefore \because I=\frac{1}{2}ln5+\frac{3}{\sqrt{5}}\times \frac{2}{3}\times \frac{\pi }{2}$$$$=\frac{1}{2}ln5+\frac{\pi }{\sqrt{5}}..$$$$$$$$$$

Answered by phanphuoc last updated on 30/Aug/21

$$=1/2{\int }_{-1}^1\frac{6x+4dx}{3+4x+3x^2}+2{\int }_{-1}^1\frac{dx}{3+4x+3x^2}$$$$=1/2ln{(1+4x+3x^2)}_{-1}^1+2/3{\int }_{-1}^1\frac{dx}{1+4/3x+x^2}$$$$=...+2/\sqrt{5}arctan{\left(\frac{x+2/3}{\sqrt{5}/3}\right)}_{-1}^1====.....$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com