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Question Number 168894 by mnjuly1970 last updated on 20/Apr/22

Commented by mahdipoor last updated on 20/Apr/22

$$\int \frac{1}{12}(\frac{3-x}{x^2+x}+\frac{x-2}{x^2+2x+4})dx=$$$$\int \frac{dx}{(x^2+x)(x^2+2x+4)}=\int \frac{d\left(lnx\right)}{(x+1)(x^2+2x+4)}=$$$$\frac{lnx}{(x+1)(x^2+2x+4)}+2\int \frac{\left(lnx\right)dx}{{(x^2+2x+4)}^2}$$$$$$

Answered by Mathspace last updated on 21/Apr/22

$$f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{lnx}{x^2+2x+a^2}dx(a>1)$$$$f^{{}^{\prime }}\left(a\right)=-2a{\int }_0^{\mathrm{\infty }}\frac{lnx}{{(x^2+2x+a^2)}^2}dx\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{lnx}{{(x^2+2x+a^2)}^2}dx=-\frac{1}{2a}{f}^{{}^{\prime }}\left(a\right)$$$$and{\int }_0^{\mathrm{\infty }}\frac{lnx}{{(x^2+2x+4)}^2}dx=-\frac{1}{4}{f}^{{}^{\prime }}\left(2\right)$$$$f\left(a\right)=-\frac{1}{2}Re\left(\mathrm{\Sigma }Res(\mathrm{\Psi },a_i)\right)$$$$\mathrm{\Psi }\left(z\right)=\frac{{ln}^{2}z}{x^2+2x+a^2}$$$${\mathrm{\Delta }}^{{}^{\prime }}=1-a^2\Rightarrow z_1=-1+i\sqrt{a^2-1}$$$$z_2=-1-i\sqrt{a^2-1}$$$$\mathrm{\Psi }\left(z\right)=\frac{{ln}^{2}z}{(z-z_1)(z-z_2)}$$$$Res(\mathrm{\Psi },z_1)=\frac{{ln}^2\left(z_1\right)}{z_1-z_2}$$$$z_1=a\Rightarrow z_1={ae}^{-iarctan\left(\sqrt{a^2-1}\right)\Rightarrow }$$$${\left({lnz}_1\right)}^2=(lna-iarctan{\left(\sqrt{a^2-1}\right)}^2$$$$=\frac{{ln}^{2}a-2ilnaarctan\left(\sqrt{a^2-1}\right)+{arctan}^2\left(\sqrt{a^2-1}\right)}{1}$$$$\Rightarrow Res(\mathrm{\Psi },z_1)=\frac{{ln}^{2}a-2ilnaarctan\left(\sqrt{a^2-1}\right)+{arcta}^2\left(\sqrt{a^2-1}\right)}{2i\sqrt{a^2-1}}$$$$...becontinued...$$

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