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Question Number 81433 by abdomathmax last updated on 13/Feb/20

$$calculate{\int }_2^{+\mathrm{\infty }}\frac{2x+3}{{(x-1)}^3{(x^2+x+1)}^2}dx$$

Answered by MJS last updated on 13/Feb/20

$$\int \frac{2x+3}{{(x-1)}^2{(x^2+x+1)}^2}=$$$$Q_1\left(x\right)=Q_2\left(x\right)=(x-1)(x^2+x+1)$$$$P_1\left(x\right)=-\frac{2}{3}{x}^2-x$$$$P_2\left(x\right)=-\frac{2}{3}x-2$$$$=-\frac{x(2x+3)}{3(x-1)(x^2+x+1)}-\frac{2}{3}\int \frac{x+3}{(x-1)(x^2+x+1)}dx=$$$$\int \frac{x+3}{(x-1)(x^2+x+1)}dx=\frac{4}{3}\int \frac{dx}{x-1}-\frac{1}{3}\int \frac{4x+5}{x^2+x+1}dx=$$$$=\frac{4}{3}\ln \mid x-1\mid -\frac{2}{3}\ln (x^2+x+1)-\frac{2\sqrt{3}}{3}\arctan \frac{\sqrt{3}(2x+1)}{3}=$$$$=\frac{2}{3}\ln \frac{{(x-1)}^2}{x^2+x+1}-\frac{2\sqrt{3}}{3}\arctan \frac{\sqrt{3}(2x+1)}{3}$$$$=-\frac{x(2x+3)}{3(x-1)(x^2+x+1)}-\frac{4}{9}\ln \frac{{(x-1)}^2}{x^2+x+1}+\frac{4\sqrt{3}}{9}\arctan \frac{\sqrt{3}(2x+1)}{3}+C$$$$\Rightarrow$$$$\mathrm{answer}\mathrm{is}\frac{2\pi \sqrt{3}}{9}+\frac{2}{3}-\frac{4}{9}\ln 7-\frac{4\sqrt{3}}{9}\arctan \frac{5\sqrt{3}}{3}$$

Commented by john santu last updated on 13/Feb/20

$$\mathrm{Ostrogradski}\mathrm{method}.\mathrm{i}\mathrm{will}\mathrm{learn}$$$$\mathrm{this}\mathrm{method}$$$$$$

Commented by abdomathmax last updated on 13/Feb/20

$$thankyousirmjs$$

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