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Question Number 132693 by liberty last updated on 15/Feb/21

$$\mathrm{I}=\int \frac{dx}{x{(x^2+1)}^3}$$

Answered by EDWIN88 last updated on 15/Feb/21

$$\mathrm{Ostrogradsky}\mathrm{again}$$$$\int \frac{dx}{x{(x^2+1)}^3}=\frac{{ax}^3+{bx}^2+cx+d}{{(x^2+1)}^2}+\int \frac{{ex}^2+fx+g}{x(x^2+1)}dx$$$$aftersolvingforcoefficient$$$$a=0,b=\frac{1}{2},c=0,d=\frac{3}{4},\mathrm{e}=0,\mathrm{f}=0\mathrm{and}\mathrm{g}=1$$$$\int \frac{dx}{x{(x^2+1)}^3}=\frac{2x^2+3}{4{(x^2+1)}^2}+\int \frac{1}{x(x^2+1)}dx$$$$=\frac{2x^2+3}{4{(x^2+1)}^2}+\int (\frac{1}{x}-\frac{x}{x^2+1})dx$$$$=\frac{2x^2+3}{4{(x^2+1)}^2}+\ln \mid x\mid -\frac{1}{2}\ln (x^2+1)+C$$$$\mathrm{Ich}\mathrm{mag}\mathrm{dieses}\mathrm{integral}.\mathrm{sch}\ddot{\mathrm{o}ner}\mathrm{ostrogradsky}$$

Answered by mathmax by abdo last updated on 16/Feb/21

$$\mathrm{I}=\int \frac{\mathrm{dx}}{\mathrm{x}{({\mathrm{x}}^2+1)}^3}\Rightarrow \mathrm{I}=\int \frac{\mathrm{dx}}{\mathrm{x}{(\mathrm{x}-\mathrm{i})}^2{(\mathrm{x}+\mathrm{i})}^2}=\int \frac{\mathrm{dx}}{\mathrm{x}{\left(\frac{\mathrm{x}-\mathrm{i}}{\mathrm{x}+\mathrm{i}}\right)}^2{(\mathrm{x}+\mathrm{i})}^4}$$$$\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\frac{\mathrm{x}-\mathrm{i}}{\mathrm{x}+\mathrm{i}}=\mathrm{t}\Rightarrow \mathrm{x}-\mathrm{i}=\mathrm{tx}+\mathrm{it}\Rightarrow (1-\mathrm{t})\mathrm{x}=\mathrm{i}(1+\mathrm{t})\Rightarrow$$$$\mathrm{x}=\frac{\mathrm{i}(1+\mathrm{t})}{1-\mathrm{t}}\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{i}\frac{(1-\mathrm{t})-(1+\mathrm{t})(-1)}{{(1-\mathrm{t})}^2}=\frac{2\mathrm{i}}{{(1-\mathrm{t})}^2}$$$$\mathrm{x}+\mathrm{i}=\mathrm{i}(\frac{1+\mathrm{t}}{1-\mathrm{t}}+1)=\mathrm{i}\left(\frac{2}{1-\mathrm{t}}\right)=\frac{2\mathrm{i}}{1-\mathrm{t}}\Rightarrow$$$$\mathrm{I}=\int \frac{1}{\mathrm{i}\frac{1+\mathrm{t}}{1-\mathrm{t}}.{\mathrm{t}}^2\frac{{\left(2\mathrm{i}\right)}^4}{{(1-\mathrm{t})}^4}}\times \frac{2\mathrm{i}}{{(1-\mathrm{t})}^2}=\frac{2}{{\left(2\mathrm{i}\right)}^4}\int \frac{{(1-\mathrm{t})}^5}{(1+\mathrm{t}){\mathrm{t}}^2{(1-\mathrm{t})}^2}\mathrm{dt}$$$$=\frac{1}{8}\int \frac{{(1-\mathrm{t})}^3}{{\mathrm{t}}^2(1+\mathrm{t})}\mathrm{dt}=\frac{1}{8}\int \frac{1-3\mathrm{t}+{3\mathrm{t}}^2-{\mathrm{t}}^3}{{\mathrm{t}}^2(\mathrm{t}+1)}\mathrm{dt}$$$$=>8\mathrm{I}=\int \frac{\mathrm{dt}}{{\mathrm{t}}^2(\mathrm{t}+1)}-\int \frac{\mathrm{dt}}{\mathrm{t}(\mathrm{t}+1)}+3\int \frac{\mathrm{dt}}{\mathrm{t}+1}-\int \frac{\mathrm{t}}{\mathrm{t}+1}\mathrm{dt}$$$$\mathrm{and}\mathrm{those}\mathrm{integrals}\mathrm{are}\mathrm{simples}\mathrm{to}\mathrm{calculate}....$$

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