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Question Number 129635 by pticantor last updated on 17/Jan/21

$$$$$$$$$$\int \frac{\mathit{d}\mathit{x}}{{({\mathit{x}}^2+1)}^2}=?$$$$\underset{\mathit{k}=1}{\sum ^{\mathit{n}}}\frac{1}{\mathit{k}(\mathit{k}+1)(2\mathit{k}+1)}=?$$

Commented by liberty last updated on 17/Jan/21

$$\frac{1}{\mathrm{k}(\mathrm{k}+1)(2\mathrm{k}+1)}=\frac{\mathrm{a}}{\mathrm{k}}+\frac{\mathrm{b}}{\mathrm{k}+1}+\frac{\mathrm{c}}{2\mathrm{k}+1}$$$$\mathrm{a}={\left[\frac{1}{(\mathrm{k}+1)(2\mathrm{k}+1)}\right]}_{\mathrm{k}=0}=1$$$$\mathrm{b}={\left[\frac{1}{\mathrm{k}(2\mathrm{k}+1)}\right]}_{\mathrm{k}=-1}=1$$$$\mathrm{c}={\left[\frac{1}{\mathrm{k}(\mathrm{k}+1)}\right]}_{\mathrm{k}=-\frac{1}{2}}=-4$$$$$$

Answered by Dwaipayan Shikari last updated on 17/Jan/21

$$\int \frac{dx}{{(x^2+1)}^2}x=tan\theta \Rightarrow 1={sec}^2\theta \frac{d\theta }{dx}$$$$=\int \frac{{sec}^2\theta }{{sec}^4\theta }d\theta \Rightarrow \int {cos}^2\theta d\theta =\frac{\theta }{2}+\frac{sin2\theta }{4}=\frac{{tan}^{-1}x}{2}+\frac{1}{2}\left(sin\theta cos\theta \right)$$$$=\frac{{tan}^{-1}x}{2}+\frac{x}{2(1+x^2)}+C$$

Answered by mathmax by abdo last updated on 17/Jan/21

$$\mathrm{I}=\int \frac{\mathrm{dx}}{{({\mathrm{x}}^2+1)}^2}\Rightarrow \mathrm{I}{=}_{\mathrm{x}=\tan \theta }\int \frac{(1+{\tan }^2\theta )}{{(1+{\tan }^2\theta )}^2}\mathrm{d}\theta =\int \frac{\mathrm{d}\theta }{1+{\tan }^2\theta }$$$$=\int {\cos }^2\theta \mathrm{d}\theta =\int \frac{1+\cos \left(2\theta \right)}{2}\mathrm{d}\theta =\frac{\theta }{2}+\frac{1}{4}\sin \left(2\theta \right)+\mathrm{C}$$$$=\frac{\mathrm{arctanx}}{2}+\frac{1}{4}\times \frac{2\tan \theta }{1+{\tan }^2\theta }+\mathrm{C}$$$$=\frac{\mathrm{arctanx}}{2}+\frac{\mathrm{x}}{2(1+{\mathrm{x}}^2)}+\mathrm{C}$$

Answered by mathmax by abdo last updated on 17/Jan/21

$$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}(\mathrm{x}+1)(2\mathrm{x}+1)}\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{x}+1}+\frac{\mathrm{c}}{2\mathrm{x}+1}$$$$\mathrm{a}=1,\mathrm{b}=1,\mathrm{c}=\frac{1}{(-\frac{1}{2})\left(\frac{1}{2}\right)}=-4\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}+1}-\frac{4}{2\mathrm{x}+1}\Rightarrow$$$${\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}(\mathrm{k}+1)(2\mathrm{k}+1)}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}+\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+1}-4\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{2\mathrm{k}+1}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}={\mathrm{H}}_{\mathrm{n}},\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+1}=\sum _{\mathrm{k}=2}^{\mathrm{n}+1}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}+1}-1$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{2\mathrm{k}+1}=\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2\mathrm{n}+1}$$$$=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{2\mathrm{n}}+\frac{1}{2\mathrm{n}+1}-1-\frac{1}{2}(1+\frac{1}{2}+...+\frac{1}{\mathrm{n}})$$$$={\mathrm{H}}_{2\mathrm{n}+1}-\frac{1}{2}{\mathrm{H}}_{\mathrm{n}}-1\Rightarrow$$$${\mathrm{S}}_{\mathrm{n}}={\mathrm{H}}_{\mathrm{n}}+{\mathrm{H}}_{\mathrm{n}+1}-1-4({\mathrm{H}}_{2\mathrm{n}+1}-\frac{1}{2}{\mathrm{H}}_{\mathrm{n}}-1)$$$$={\mathrm{H}}_{\mathrm{n}}+{\mathrm{H}}_{\mathrm{n}+1}-1-{4\mathrm{H}}_{2\mathrm{n}+1}+{2\mathrm{H}}_{\mathrm{n}}+4$$$$={2\mathrm{H}}_{\mathrm{n}}+\frac{1}{\mathrm{n}+1}-{4\mathrm{H}}_{2\mathrm{n}+1}+3+{2\mathrm{H}}_{\mathrm{n}}$$$$=4{\mathrm{H}}_{\mathrm{n}}-{4\mathrm{H}}_{2\mathrm{n}+1}+3$$

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