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Question Number 147683 by mathmax by abdo last updated on 22/Jul/21

$$\mathrm{let}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{{(\mathrm{x}+1)}^5{(2\mathrm{x}-3)}^4}$$$$1)\mathrm{find}\int \mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}$$$$2)\mathrm{en}\mathrm{deduire}\mathrm{la}\mathrm{decomposition}\mathrm{de}\mathrm{F}\mathrm{en}\mathrm{element}\mathrm{simples}$$

Answered by mathmax by abdo last updated on 23/Jul/21

$$1)\mathrm{I}=\int \frac{\mathrm{dx}}{{(\mathrm{x}+1)}^5{(2\mathrm{x}-3)}^4}\Rightarrow \mathrm{I}=\int \frac{\mathrm{dx}}{{\left(\frac{\mathrm{x}+1}{2\mathrm{x}-3}\right)}^5{(2\mathrm{x}-3)}^9}$$$$\mathrm{changement}\frac{\mathrm{x}+1}{2\mathrm{x}-3}=\mathrm{t}\mathrm{give}\mathrm{x}+1=2\mathrm{tx}-3\mathrm{t}\Rightarrow (1-2\mathrm{t})\mathrm{x}=-1-3\mathrm{t}\Rightarrow$$$$\mathrm{x}=\frac{3\mathrm{t}+1}{2\mathrm{t}-1}\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{3(2\mathrm{t}-1)-2(3\mathrm{t}+1)}{{(2\mathrm{t}-1)}^2}=\frac{-5}{{(2\mathrm{t}-1)}^2}$$$$2\mathrm{x}-3=\frac{6\mathrm{t}+2}{2\mathrm{t}-1}-3=\frac{6\mathrm{t}+2-6\mathrm{t}+3}{2\mathrm{t}-1}=\frac{5}{2\mathrm{t}-1}\Rightarrow$$$$\mathrm{I}\left(\mathrm{x}\right)=\int \frac{-5}{{(2\mathrm{t}-1)}^2{\mathrm{t}}^5{\left(\frac{5}{2\mathrm{t}-1}\right)}^9}\mathrm{dt}$$$$=-\frac{1}{5^8}\int \frac{{(2\mathrm{t}-1)}^9}{{(2\mathrm{t}-1)}^2{\mathrm{t}}^5}\mathrm{dt}=-\frac{1}{5^8}\int \frac{{(2\mathrm{t}-1)}^7}{{\mathrm{t}}^5}\mathrm{dt}$$$$=-\frac{1}{5^8}\int \frac{\sum _{\mathrm{k}=0}^7{\mathrm{C}}_7^{\mathrm{k}}{\left(2\mathrm{t}\right)}^{\mathrm{k}}{(-1)}^{7-\mathrm{k}}}{{\mathrm{t}}^5}\mathrm{dt}$$$$=\frac{1}{5^8}\int \sum _{\mathrm{k}=0}^7{\mathrm{C}}_7^{\mathrm{k}}{(-2)}^{\mathrm{k}}{\mathrm{t}}^{\mathrm{k}-5}\mathrm{dt}$$$$=\frac{1}{5^8}\sum _{\mathrm{k}=0}^7{(-2)}^{\mathrm{k}}{\mathrm{C}}_7^{\mathrm{k}}\int {\mathrm{t}}^{\mathrm{k}-5}\mathrm{dt}$$$$=\frac{1}{5^8}\sum _{\mathrm{k}=0\mathrm{and}\mathrm{k}\ne 4}^7{(-2)}^{\mathrm{k}}{\mathrm{C}}_7^{\mathrm{k}}\times \frac{1}{\mathrm{k}-4}{\mathrm{t}}^{\mathrm{k}-4}+\frac{1}{5^8}{(-2)}^4{\mathrm{C}}_7^4\log \mid \mathrm{t}\mid +\mathrm{C}$$$$\mathrm{I}\left(\mathrm{x}\right)=\frac{1}{5^8}\sum _{\mathrm{k}=0\mathrm{and}\mathrm{k}\ne 4}^7\frac{{(-2)}^{\mathrm{k}}}{\mathrm{k}-4}{\mathrm{C}}_7^{\mathrm{k}}{\left(\frac{\mathrm{x}+1}{2\mathrm{x}-3}\right)}^{\mathrm{k}-4}+\frac{2^4}{5^8}{\mathrm{C}}_7^4\log \mid \frac{\mathrm{x}+1}{2\mathrm{x}-3}\mid +\mathrm{C}$$

Commented by mathmax by abdo last updated on 23/Jul/21

$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{dI}}{\mathrm{dx}}\left(\mathrm{x}\right)$$$$=\frac{1}{5^8}\sum _{\mathrm{k}=0\mathrm{and}\mathrm{k}\ne 4}^7{(-2)}^{\mathrm{k}}{\mathrm{C}}_7^{\mathrm{k}}{\left(\frac{\mathrm{x}+1}{2\mathrm{x}-3}\right)}^{{}^{\prime }}{\left(\frac{\mathrm{x}+1}{2\mathrm{x}-3}\right)}^{\mathrm{k}-5}+\frac{2^4}{5^8}{\mathrm{C}}_7^4\frac{{\left(\frac{\mathrm{x}+1}{2\mathrm{x}-3}\right)}^{{}^{\prime }}}{\frac{\mathrm{x}+1}{2\mathrm{x}-3}}$$$$=-\frac{1}{5^7}\sum _{\mathrm{k}=0\mathrm{andk}\ne 4}^7{(-2)}^{\mathrm{k}}{\mathrm{C}}_7^{\mathrm{k}}\frac{{(\mathrm{x}+1)}^{\mathrm{k}-5}}{{(2\mathrm{x}-3)}^{\mathrm{k}-3}}-\frac{2^4}{5^7}{\mathrm{C}}_7^4\frac{2\mathrm{x}-3}{\mathrm{x}+1}\times \frac{1}{{(2\mathrm{x}-3)}^2}$$$$....\mathrm{be}\mathrm{continued}....$$

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