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Question Number 183382 by mathlove last updated on 25/Dec/22

Commented by CElcedricjunior last updated on 25/Dec/22

$$\int \frac{\mathit{x}+1}{2{\mathit{x}}^2+6\mathit{x}+9}\mathit{d}\mathit{x}=\frac{1}{4}\int \frac{4\mathit{x}+6}{2{\mathit{x}}^2+6\mathit{x}+9}\mathit{d}\mathit{x}-\frac{1}{4}\int \frac{-2}{2{\mathit{x}}^2+6\mathit{x}+9}\mathit{d}\mathit{x}$$$$=\frac{1}{4}\mathit{l}\mathit{n}\mid 2{\mathit{x}}^2+6\mathit{x}+9\mid -\frac{1}{4}\int \frac{dx}{{(\mathit{x}+\frac{3}{2})}^2+\frac{9}{4}}\mathit{t}=\frac{2\mathit{x}+3}{3}$$$$=\frac{1}{4}\mathbf{ln}\mid 2{\mathit{x}}^2+6\mathit{x}+9\mid -\frac{1}{6}\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\frac{2\mathit{x}+3}{3}\right)+\mathit{c}$$

Commented by Acem last updated on 25/Dec/22

$$Salut,vousinsistezpourmettretoutevotre$$$$solutionsencommentaires!$$

Commented by mr W last updated on 26/Dec/22

$$To\mathit{C}\mathit{E}\mathit{l}\mathit{c}\mathit{e}\mathit{d}\mathit{r}\mathit{i}\mathit{c}\mathit{j}\mathit{u}\mathit{n}\mathit{i}\mathit{o}\mathit{r}sir:$$$$lastwarning!$$$$Whenyouinsistondisobeyingthe$$$$rulesoftheforumandignorethe$$$$kindadvisesofotherpeople,thenthis$$$$forumisnotasuitableplaceforyou,$$$$inotherwords,youarenotwelcome$$$$here,and{I}^{\prime }llasktheadministrator$$$$tobanyoufromtheforum!$$$$Imageyouarewaitingforthebus$$$$atabusstopinaqueue,thenaguy$$$$comesandpusheshimselfbeforeall$$$$otherpeopleinthequeue.Your$$$$currentbehaviorisexactlythesame$$$$asthisguy.Sopleaseleavehereor$$$$obeytheruleshere!$$$$ifyouhavespecialreasonforyour$$$$doing,pleaseletusknowandspeak$$$$itout\left(inFrenchwhenyoulike\right).Just$$$$ignoringotherpeoplewithoutsaying$$$$anythingisnotpolite.$$

Commented by CElcedricjunior last updated on 26/Dec/22

$$i^{\prime }mverysor$$

Answered by Acem last updated on 25/Dec/22

$$I=\int \frac{x+1}{2x^2+6x+9}dx=\frac{1}{2}\int \frac{x+1}{{(x+\frac{3}{2})}^2+\frac{9}{4}}dx$$$$t=x+\frac{3}{2}...i$$$$I=\frac{1}{4}\int \frac{2t}{t^2+\frac{9}{4}}dt-\frac{1}{6}\int \frac{d\left(\frac{2}{3}t\right)}{{\left(\frac{2}{3}t\right)}^2+1}$$$$I=\frac{1}{4}\ln (x^2+3x+\frac{9}{2})-\frac{1}{6}\arctan (\frac{2}{3}x+1)+c$$$$$$

Answered by Tony6400 last updated on 25/Dec/22

$$I=\int \frac{x}{2x^2+6x+9}dx+\int \frac{1}{2x^2+6x+9}dx=A+B$$$$NowA=\int \frac{x}{2x^2+6x+9}dx=\int \frac{x}{2(x^2+3x+\frac{9}{2})}dx=\frac{1}{2}\int \frac{x}{x^2+3x+{\left(\frac{3}{2}\right)}^2+\frac{9}{2}-{\left(\frac{3}{2}\right)}^2}dx$$$$\Rightarrow A=\frac{1}{2}\int \frac{x}{{(x+\frac{3}{2})}^2+{\left(\frac{3}{2}\right)}^2}dx.Letx+\frac{3}{2}=\frac{3}{2}tan\theta \therefore x=\frac{3}{2}(tan\theta -1)$$$$\Rightarrow \frac{dx}{d\theta }=\frac{3}{2}{sec}^2\theta \therefore dx=\frac{3}{2}{sec}^2\theta d\theta$$$$\Rightarrow A=\frac{1}{2}\int \frac{\frac{3}{2}(tan\theta -1)}{{\left(\frac{3}{2}tan\theta \right)}^2+\frac{9}{4}}.\frac{3}{2}{sec}^2\theta d\theta$$$$\therefore A=\frac{1}{2}\int \frac{\frac{9}{4}{sec}^2\theta (tan\theta -1)d\theta }{\frac{9}{4}({tan}^2\theta +1)}=\frac{1}{2}\int (tan\theta -1)d\theta =\frac{1}{2}[-ln\left(cos\theta \right)-\theta ].Fromx+\frac{3}{2}=\frac{3}{2}tan\theta \Rightarrow tan\theta =\frac{2x+3}{3}=1+\frac{2}{3}x\therefore cos\theta =\frac{3}{\sqrt{{(2x+3)}^2+9}}$$$$And\theta =arctan(1+\frac{2}{3}x)\therefore A=-\frac{1}{2}[ln(\frac{3}{\sqrt{{(2x+3)}^2+9}}+arctan(\frac{2}{3}x+1)]+C_1$$$$MoreoverB=\int \frac{1}{2x^2+6x+9}dx=\frac{1}{2}\int \frac{1}{{(x+\frac{3}{2})}^2+{\left(\frac{3}{2}\right)}^2}dx=\frac{1}{2}.\frac{2}{3}arctan(1+\frac{2}{3}x)+C_2$$$$\Rightarrow B=\frac{arctan(1+\frac{2}{3}x)}{3}+C_2$$$$\Rightarrow I=A+B=\int \frac{x+1}{2x^2+6x+9}dx=-\frac{1}{2}[ln\left(\frac{3}{\sqrt{{(2x+3)}^2+9}}\right)+arctan(\frac{2}{3}x+1)]+\frac{arctan(1+\frac{2}{3}x)}{3}+K,whereK=C_1+C_2$$$$I=-\frac{1}{2}ln\left(\frac{3}{\sqrt{{(2x+3)}^2+9}}\right)-\frac{arctan(\frac{2}{3}x+1)}{2}+\frac{arctan(1+\frac{2}{3}x)}{3}+K=-\frac{1}{2}ln\left(\frac{3}{\sqrt{{(2x+3)}^2+9}}\right)-\frac{arctan(\frac{2}{3}x+1)}{6}$$$$$$$$$$$$$$

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