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Question Number 97153 by eidmarie last updated on 06/Jun/20

Answered by MJS last updated on 06/Jun/20

$$-\int \frac{x^3-x^2-2x-13}{(x+1)(x^2-x+3)}dx=$$$$=-\int dx+\frac{13}{5}\int \frac{dx}{x+1}-\frac{1}{5}\int \frac{8x-41}{x^2-x+3}dx=$$$$=-x+\frac{13}{5}\ln \mid x+1\mid -\frac{4}{5}\int \frac{2x-1}{x^2-x+3}dx+\frac{37}{5}\int \frac{dx}{x^2-x+3}=$$$$=-x+\frac{13}{5}\ln \mid x+1\mid -\frac{4}{5}\ln (x^2-x+3)+\frac{74\sqrt{11}}{55}\arctan \frac{\sqrt{11}(2x-1)}{11}+C$$$$\mathrm{now}\mathrm{insert}\mathrm{borders}$$

Answered by mathmax by abdo last updated on 06/Jun/20

$$\mathrm{A}={\int }_{-3}^0\frac{-{\mathrm{x}}^3+{\mathrm{x}}^2+2\mathrm{x}+13}{{\mathrm{x}}^3+2\mathrm{x}+3}\mathrm{dx}\mathrm{changement}\mathrm{x}=-\mathrm{t}\mathrm{give}$$$$\mathrm{A}={\int }_0^3\frac{{\mathrm{t}}^3+{\mathrm{t}}^2-2\mathrm{t}+13}{-{\mathrm{t}}^3-2\mathrm{t}+3}\mathrm{dt}=-{\int }_0^3\frac{{\mathrm{t}}^3+{\mathrm{t}}^2-2\mathrm{t}+13}{{\mathrm{t}}^3+2\mathrm{t}-3}\mathrm{dt}$$$$\mathrm{A}=-{\int }_0^3\frac{{\mathrm{t}}^3+2\mathrm{t}-3-2\mathrm{t}+3+{\mathrm{t}}^2-2\mathrm{t}+13}{{\mathrm{t}}^3+2\mathrm{t}-3}\mathrm{dt}$$$$=-3-{\int }_0^3\frac{{\mathrm{t}}^2-4\mathrm{t}+16}{{\mathrm{t}}^3+2\mathrm{t}-3}\mathrm{dt}\mathrm{we}\mathrm{have}1\mathrm{is}\mathrm{root}\mathrm{of}\mathrm{p}\left(\mathrm{t}\right)={\mathrm{t}}^3+2\mathrm{t}-3$$$${\mathrm{t}}^3+2\mathrm{t}-3={\mathrm{t}}^3-1+2\mathrm{t}-2=(\mathrm{t}-1)({\mathrm{t}}^2+\mathrm{t}+1)+2(\mathrm{t}-1)$$$$=(\mathrm{t}-1)({\mathrm{t}}^2+\mathrm{t}+3)\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{t}\right)=\frac{{\mathrm{t}}^2-4\mathrm{t}+16}{(\mathrm{t}-1)({\mathrm{t}}^2+\mathrm{t}+3)}$$$$=\frac{\mathrm{a}}{\mathrm{t}-1}+\frac{\mathrm{bt}+\mathrm{c}}{{\mathrm{t}}^2+\mathrm{t}+3}$$$$\mathrm{a}=\frac{13}{5},{\lim }_{\mathrm{t}\to +\mathrm{\infty }}\mathrm{tF}\left(\mathrm{t}\right)=1=\mathrm{a}+\mathrm{b}\Rightarrow \mathrm{b}=1-\frac{13}{5}=-\frac{8}{5}$$$$\mathrm{F}\left(0\right)=-\frac{16}{3}=-\mathrm{a}+\mathrm{c}\Rightarrow \mathrm{c}=\mathrm{a}-\frac{16}{3}=\frac{13}{5}-\frac{16}{3}=\frac{39-80}{15}=-\frac{41}{15}\Rightarrow$$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{13}{5(\mathrm{t}-1)}+\frac{-\frac{8}{5}\mathrm{t}-\frac{41}{15}}{{\mathrm{t}}^2+\mathrm{t}+3}\Rightarrow \int \mathrm{F}\left(\mathrm{t}\right)\mathrm{dt}=\frac{13}{5}\ln \mid \mathrm{t}-1\mid -\frac{8}{10}\int \frac{2\mathrm{t}+1-1}{{\mathrm{t}}^2+\mathrm{t}+3}\mathrm{dt}$$$$-\frac{41}{15}\int \frac{\mathrm{dt}}{{\mathrm{t}}^2+\mathrm{t}+3}\mathrm{now}\mathrm{its}\mathrm{eazy}\mathrm{to}\mathrm{solve}....$$$$$$

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