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Question Number 95586 by turbo msup by abdo last updated on 26/May/20

$$find\int \frac{dx}{x^n{(x+1)}^m}$$$$mandnintegr$$

Commented by Tony Lin last updated on 26/May/20

$$letu=\frac{x}{x+1},dx={(x+1)}^{2}du$$$$x=\frac{u}{1-u}$$$$\int \frac{1}{{\left(\frac{u}{1-u}\right)}^n{\left(\frac{1}{1-u}\right)}^{m-2}}du$$$$=\int \frac{{(1-\frac{x}{x+1})}^{n+m-2}}{{\left(\frac{x}{x+1}\right)}^n}d\left(\frac{x}{x+1}\right)$$$$=C_{n-1}^{n+m-2}{(-1)}^{n-1}ln\mid \frac{x}{x+1}\mid$$$$+\underset{k=0}{\sum ^{n-2}}{C}_k^{n+m-2}{(-1)}^{k+1}\frac{1}{(n+k-1){\left(\frac{x}{x+1}\right)}^{u+k-1}}$$$$+\underset{k=0}{\sum ^{m-2}}{C}_{n+k}^{n+m-2}{(-1)}^{n+k}\frac{{\left(\frac{x}{x+1}\right)}^{k+1}}{k+1}$$$$+c$$

Answered by mathmax by abdo last updated on 26/May/20

$$\mathrm{I}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^{\mathrm{n}}{(\mathrm{x}+1)}^{\mathrm{m}}}\Rightarrow \mathrm{I}=\int \frac{\mathrm{dx}}{{\left(\frac{\mathrm{x}}{\mathrm{x}+1}\right)}^{\mathrm{n}}{(\mathrm{x}+1)}^{\mathrm{m}+\mathrm{n}}}$$$$\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\frac{\mathrm{x}}{\mathrm{x}+1}=\mathrm{t}\Rightarrow \mathrm{x}=\mathrm{tx}+\mathrm{t}\Rightarrow (1-\mathrm{t})\mathrm{x}=\mathrm{t}$$$$\Rightarrow \mathrm{x}=\frac{\mathrm{t}}{1-\mathrm{t}}\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1-\mathrm{t}-\mathrm{t}(-1)}{{(1-\mathrm{t})}^2}=\frac{1}{{(1-\mathrm{t})}^2}\mathrm{and}\mathrm{x}+1=\frac{\mathrm{t}}{1-\mathrm{t}}+1=\frac{1}{1-\mathrm{t}}$$$$\mathrm{I}=\int \frac{\mathrm{dt}}{{(1-\mathrm{t})}^2{\mathrm{t}}^{\mathrm{n}}{\left(\frac{1}{1-\mathrm{t}}\right)}^{\mathrm{m}+\mathrm{n}}}=\int \frac{{(1-\mathrm{t})}^{\mathrm{m}+\mathrm{n}}}{{(1-\mathrm{t})}^2{\mathrm{t}}^{\mathrm{n}}}\mathrm{dt}$$$$=\int \frac{{(1-\mathrm{t})}^{\mathrm{m}+\mathrm{n}-2}}{{\mathrm{t}}^{\mathrm{n}}}\mathrm{dt}={(-1)}^{\mathrm{m}+\mathrm{n}-2}\int \frac{{(\mathrm{t}-1)}^{\mathrm{m}+\mathrm{n}-2}}{{\mathrm{t}}^{\mathrm{n}}}\mathrm{dt}$$$$={(-1)}^{\mathrm{m}+\mathrm{n}-2}\int \frac{\sum _{\mathrm{k}=0}^{\mathrm{m}+\mathrm{n}-2}{\mathrm{C}}_{\mathrm{m}+\mathrm{n}-2}^{\mathrm{k}}{\mathrm{t}}^{\mathrm{k}}}{{\mathrm{t}}^{\mathrm{n}}}\mathrm{dt}$$$$={(-1)}^{\mathrm{m}+\mathrm{n}-2}\sum _{\mathrm{k}=0}^{\mathrm{m}+\mathrm{n}-2}{\mathrm{C}}_{\mathrm{m}+\mathrm{n}-2}^{\mathrm{k}}\int {\mathrm{t}}^{\mathrm{k}-\mathrm{n}}\mathrm{dt}$$$$={(-1)}^{\mathrm{m}+\mathrm{n}-2}\sum _{\mathrm{k}=0\mathrm{and}\mathrm{k}\ne \mathrm{n}-1}^{\mathrm{m}+\mathrm{n}-2}{\mathrm{C}}_{\mathrm{m}+\mathrm{n}-2}^{\mathrm{k}}\frac{1}{\mathrm{k}-\mathrm{n}+1}{\mathrm{t}}^{\mathrm{k}-\mathrm{n}+1}$$$$+{(-1)}^{\mathrm{m}+\mathrm{n}-2}{\mathrm{C}}_{\mathrm{m}+\mathrm{n}-2}^{\mathrm{n}-1}\ln \mid \mathrm{t}\mid +\mathrm{C}$$$$\Rightarrow \mathrm{I}={(-1)}^{\mathrm{m}+\mathrm{n}-2}\sum _{\mathrm{k}=0\mathrm{and}\mathrm{k}\ne \mathrm{n}-1}^{\mathrm{m}+\mathrm{n}-2}{\mathrm{C}}_{\mathrm{m}+\mathrm{n}-2}^{\mathrm{k}}\frac{1}{\mathrm{k}-\mathrm{n}+1}{\left(\frac{\mathrm{x}}{x+1}\right)}^{k-n+1}$$$$+{(-1)}^{m+n-2}{C}_{\mathrm{m}+\mathrm{n}-2}^{\mathrm{n}-1}\ln \mid \frac{\mathrm{x}}{\mathrm{x}+1}\mid +\mathrm{C}$$

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