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Question Number 118010 by TANMAY PANACEA last updated on 14/Oct/20

$$\int \frac{dx}{x^4+x^2+1}$$

Commented by mmmmmm1 last updated on 14/Oct/20

$$\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left[\frac{\mathit{f}\left(\mathit{x}\right)}{\mathit{g}\left(\mathit{x}\right)}\right]=\frac{\mathit{g}\left(\mathit{x}\right)\cdot \frac{\mathit{d}}{\mathit{d}\mathit{x}}\left[\mathit{f}\left(\mathit{x}\right)\right]-\mathit{f}\left(\mathit{x}\right)\cdot \frac{\mathit{d}}{\mathit{d}\mathit{x}}\left[\mathit{g}\left(\mathit{x}\right)\right]}{\mathit{g}{\left(\mathit{x}\right)}^2}$$$$\frac{\mathit{d}-3{\mathit{x}}^4\mathit{d}-{\mathit{x}}^2\mathit{d}}{{({\mathit{x}}^4+{\mathit{x}}^2+1)}^2}$$

Answered by MJS_new last updated on 14/Oct/20

$$\int \frac{dx}{x^4+x^2+1}=$$$$=\frac{1}{2}\int \frac{x+1}{x^2+x+1}dx-\frac{1}{2}\int \frac{x-1}{x^2-x+1}dx=$$$$=\frac{1}{4}\int \frac{dx}{x^2+x+1}+\frac{1}{4}\int \frac{2x+1}{x^2+x+1}dx+$$$$+\frac{1}{4}\int \frac{dx}{x^2-x+1}-\frac{1}{4}\int \frac{2x-1}{x^2-x+1}dx=$$$$=\frac{\sqrt{3}}{6}\arctan \frac{\sqrt{3}(2x+1)}{3}+\frac{1}{4}\ln (x^2+x+1)+$$$$+\frac{\sqrt{3}}{6}\arctan \frac{\sqrt{3}(2x-1)}{3}-\frac{1}{4}\ln (x^2-x+1)=$$$$=\frac{\sqrt{3}}{6}(\arctan \frac{\sqrt{3}(2x+1)}{3}+\arctan \frac{\sqrt{3}(2x-1)}{3})+$$$$+\frac{1}{4}\ln \frac{x^2+x+1}{x^2-x+1}+C$$

Answered by TANMAY PANACEA last updated on 14/Oct/20

$$\int \frac{\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1}dx$$$$\frac{1}{2}\int \frac{(1+\frac{1}{x^2})-(1-\frac{1}{x^2})}{x^2+\frac{1}{x^2}+1}dx$$$$\frac{1}{2}\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+3}+\frac{1}{2}\int \frac{d(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2-1}$$$$nowuseformula$$$$\int \frac{dy}{y^2+a^2}and\int \frac{dy}{y^2-a^2}$$

Answered by mathmax by abdo last updated on 14/Oct/20

$$\mathrm{let}\mathrm{A}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^4+{\mathrm{x}}^2+1}\Rightarrow \mathrm{A}=\int \frac{\frac{1}{{\mathrm{x}}^2}}{{\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}+1}\mathrm{dx}$$$$=\frac{1}{2}\int \frac{1+\frac{1}{{\mathrm{x}}^2}-(1-\frac{1}{{\mathrm{x}}^2})}{{\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}+1}\mathrm{dx}=\frac{1}{2}\int \frac{1+\frac{1}{{\mathrm{x}}^2}}{{(\mathrm{x}-\frac{1}{\mathrm{x}})}^2+3}\mathrm{dx}(\to \mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{u})-\frac{1}{2}\int \frac{1-\frac{1}{\mathrm{x}}}{{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-1}(\to \mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{v})$$$$=\frac{1}{2}\int \frac{\mathrm{du}}{{\mathrm{u}}^2+3}-\frac{1}{2}\int \frac{\mathrm{dv}}{{\mathrm{v}}^2-1}\mathrm{but}\int \frac{\mathrm{du}}{{\mathrm{u}}^2+3}{=}_{\mathrm{u}=\sqrt{3}\mathrm{z}}\int \frac{\sqrt{3}\mathrm{dz}}{3({\mathrm{z}}^2+1)}$$$$=\frac{1}{\sqrt{3}}\arctan \left(\frac{\mathrm{u}}{\sqrt{3}}\right)+{\mathrm{c}}_1$$$$\int \frac{\mathrm{dv}}{{\mathrm{v}}^2-1}=\frac{1}{2}\int (\frac{1}{\mathrm{v}-1}-\frac{1}{\mathrm{v}+1})\mathrm{dv}=\frac{1}{2}\ln \mid \frac{\mathrm{v}-1}{\mathrm{v}+1}\mid +{\mathrm{c}}_2\Rightarrow$$$$\mathrm{A}=\frac{1}{2\sqrt{3}}\arctan \left(\frac{\mathrm{x}-\frac{1}{\mathrm{x}}}{\sqrt{3}}\right)-\frac{1}{4}\ln \mid \frac{\mathrm{x}+\frac{1}{\mathrm{x}}-1}{\mathrm{x}+\frac{1}{\mathrm{x}}+1}\mid +\mathrm{C}$$

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