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Question Number 42085 by maxmathsup by imad last updated on 17/Aug/18

$$calculate{\int }_1^{+\mathrm{\infty }}\frac{2x+1}{3+{(x+1)}^3}dx$$

Commented by maxmathsup by imad last updated on 18/Aug/18

$$letA={\int }_1^{+\mathrm{\infty }}\frac{2x+1}{3+{(x+1)}^3}dxchangementx+1=tgive$$$$A={\int }_2^{+\mathrm{\infty }}\frac{2(t-1)+1}{3+t^3}dt={\int }_2^{+\mathrm{\infty }}\frac{2t-1}{t^3+3}dtalsochangement{}^3\sqrt{3}u=tgive$$$$A={\int }_{\frac{2}{3_{\sqrt{3}}}}^{+\mathrm{\infty }}\frac{2\left({}^3\sqrt{3}\right)u-1}{3(1+u^3)}{}^3\sqrt{3}dulet\alpha =\frac{2}{\left({}^3\sqrt{3}\right)}$$$$=\frac{2}{3}{\left({}^3\sqrt{3}\right)}^2{\int }_{\alpha }^{+\mathrm{\infty }}\frac{u}{u^3+1}{-}^3\sqrt{3}{\int }_{\alpha }^{+\mathrm{\infty }}\frac{du}{u^3+1}$$$$=\frac{2}{3}{\left(\frac{2}{\alpha }\right)}^2{\int }_{\alpha }^{+\mathrm{\infty }}\frac{u}{u^3+1}du-\frac{2}{\alpha }{\int }_{\alpha }^{+\mathrm{\infty }}\frac{du}{u^3+1}$$$$={\int }_{\alpha }^{+\mathrm{\infty }}\frac{\frac{8}{3{\alpha }^2}u-\frac{2}{\alpha }}{u^3+1}du=\frac{2}{\alpha }{\int }_{\alpha }^{+\mathrm{\infty }}\frac{\frac{4}{3\alpha }u-1}{u^3+1}du$$$$=\frac{2}{3{\alpha }^2}{\int }_{\alpha }^{+\mathrm{\infty }}\frac{4u-3\alpha }{u^3+1}duletdecomposeF\left(u\right)=\frac{4u-3\alpha }{u^3+1}$$$$F\left(u\right)=\frac{4u-3\alpha }{(u+1)(u^2-u+1)}=\frac{a}{u+1}+\frac{bu+c}{u^2-u+1}$$$$a={lim}_{u\to -1}(u+1)F\left(u\right)=\frac{-4-3\alpha }{3}$$$${lim}_{u\to +\mathrm{\infty }}uF\left(u\right)=0=a+b\Rightarrow b=\frac{4+3\alpha }{3}\Rightarrow$$$$F\left(u\right)=-\frac{3\alpha +4}{3(u+1)}+\frac{1}{3}\frac{(3\alpha +4)u+3c}{u^2-u+1}$$$$F\left(0\right)=-3\alpha =-\frac{3\alpha +4}{3}+c\Rightarrow c=-3\alpha +\frac{3\alpha +4}{3}=\frac{-6\alpha +4}{3}\Rightarrow$$$$F\left(u\right)=-\frac{3\alpha +4}{3(u+1)}+\frac{1}{3}\frac{(3\alpha +4)u-6\alpha +4}{u^2-u+1}\Rightarrow$$$${\int }_{\alpha }^{+\mathrm{\infty }}F\left(u\right)du=(-\alpha -\frac{4}{3}){\int }_{\alpha }^{+\mathrm{\infty }}\frac{du}{u+1}+\frac{3\alpha +4}{6}{\int }_{\alpha }^{+\mathrm{\infty }}\frac{2\alpha -1+1}{u^2-u+1}du$$$$+\frac{4-6\alpha }{3}{\int }_{\alpha }^{+\mathrm{\infty }}\frac{du}{u^2-u+1}$$$$={[(-\alpha -\frac{4}{3})ln\mid u+1\mid +\frac{3\alpha +4}{6}ln(u^2-u+1)]}_0^{+\mathrm{\infty }}+\frac{4-6\alpha }{3}{\int }_{\alpha }^{+\mathrm{\infty }}\frac{du}{u^2-u+1}...$$$$$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Aug/18

$$t=x+1dt=dx$$$${\int }_2^{\mathrm{\infty }}\frac{2(t-1)+1}{3+t^3}dt$$$${\int }_2^{\mathrm{\infty }}\frac{2t+1}{3+t^3}dt$$$$\frac{2t+1}{t^3+{\left(3^{\frac{1}{3}}\right)}^3}=\frac{2t+1}{(t+3^{\frac{1}{3}})(t^2-3^{\frac{1}{3}}t+3^{\frac{2}{3}})}$$$$leta=3^{\frac{1}{3}}$$$$\frac{2t+1}{(t+a)(t^2-ta+a^2)}=\frac{P}{t+a}+\frac{Qt+R}{(t^2-ta+a^2)}$$$$2t+1=P(t^2-ta+a^2)+(t+a)(Qt+R)$$$$2t+1=t^2\left(P\right)+t(-Pa)+\left({Pa}^2\right)+t^2\left(Q\right)+t\left(R\right)+t\left(Qa\right)+aR$$$$2t+1=t^2(P+Q)+t(-Pa+R+Qa)+({Pa}^2+aR)$$$$P+Q=0$$$$-Pa+Qa+R=2$$$${Pa}^2+aR=1$$$$Q=-P$$$$-Pa-Pa+\frac{1-{Pa}^2}{a}=2$$$$-2{Pa}^2+1-{Pa}^2=2a$$$$-3{Pa}^2=2a-1$$$$P=\frac{1-2a}{3a^2}Q=\frac{2a-1}{3a^2}$$$$R=\frac{1-{Pa}^2}{a}=\frac{1-\frac{1-2a}{3}}{a}=\frac{3-1+2a}{3a}=\frac{2+2a}{3a}$$$${\int }_2^{\mathrm{\infty }}\frac{P}{t+a}+\frac{Qt+R}{t^2-ta+a^2}dt$$$$P{\int }_2^{\mathrm{\infty }}\frac{dt}{t+a}+\frac{Q}{2}{\int }_2^{\mathrm{\infty }}\frac{2t-a+a}{t^2-ta+a^2}+R{\int }_2^{\mathrm{\infty }}\frac{dt}{t^2-2.t.\frac{a}{2}+\frac{a^2}{4}+\frac{3a^2}{4}}$$$$P{\int }_2^{\mathrm{\infty }}\frac{dt}{t+a}+\frac{Q}{2}{\int }_2^{\mathrm{\infty }}\frac{2t-a}{t^2-ta+a^2}dt+(\frac{Qa}{2}+R){\int }_2^{\mathrm{\infty }}\frac{dt}{{(t-\frac{a}{2})}^2+{\left(\frac{a\sqrt{3}}{2}\right)}^2}$$$$Pln\mid (t+a){\mid }_2^{\mathrm{\infty }}+\frac{Q}{2}ln\mid (t^2-ta+a^2){\mid }_2^{\mathrm{\infty }}+(\frac{Qa}{2}+R)\times \frac{2}{a\sqrt{3}}\mid {tan}^{-1}\left(\frac{t-\frac{a}{2}}{\frac{a\sqrt{3}}{2}}\right){\mid }_2^{\mathrm{\infty }}$$$$tobeclmpleted...$$$$$$$$$$

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