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Question Number 63351 by aliesam last updated on 02/Jul/19

Commented by mathmax by abdo last updated on 03/Jul/19

$$letI=\int \frac{dx}{4+{(x-3)}^2}changementx-3=2tgive$$$$I=\int \frac{2dt}{4+4t^2}=\int \frac{dt}{2(1+t^2)}=\frac{1}{2}arctan\left(t\right)+c$$$$I=\frac{1}{2}arctan\left(\frac{x-3}{2}\right)+c.$$

Commented by mathmax by abdo last updated on 03/Jul/19

$$letA=\int \frac{2{tan}^{3}x+2tanx+1}{tanx{(1+{tan}^{2}x)}^2}dxiftan{\left(x\right)}^{\left(2\right)}means\frac{d^2}{{dx}^2}tanxweget$$$${tanx}^{\left(1\right)}=1+{tan}^{2}x\Rightarrow {tanx}^{\left(2\right)}=2tanx(1+{tan}^{2}x)\Rightarrow$$$$I=\int \frac{2{tan}^{3}x+2tanx+1}{tanx{(1+2tanx(1+{tan}^{2}x))}^2}dx=\int \frac{2{tan}^{3}x+2tanx+1}{tanx{(2{tan}^{3}x+2tanx+1)}^2}dx$$$$=\int \frac{dx}{tanx(2{tan}^{3}x+2tanx+1)}changementtanx=tgive$$$$I=\int \frac{1}{t(2t^3+2t+1)}\frac{dt}{1+t^2}=\int \frac{dt}{t(1+t^2)(2t^3+2t+1)}$$$$letdecomposeF\left(t\right)=\frac{1}{t(1+t^2)(2t^3+2t+1)}\Rightarrow$$$$F\left(t\right)=\frac{a}{t}+\frac{bt+c}{t^2+1}+\frac{{dt}^2+et+f}{2t^3+2t+1}\Rightarrow$$$$\int F\left(t\right)dt=\int \frac{adt}{t}+\int \frac{bt+c}{t^2+1}+\int \frac{{dt}^2+et+f}{2t^3+2t+1}resttocalculatethecoefficient{a}_i$$$$becontinued...$$$$$$

Answered by Rio Michael last updated on 02/Jul/19

$$1.\int \frac{dx}{4+{(x-3)}^2}=\int \frac{1}{x^2-6x+13}dx$$$$\Rightarrow ln\mid x^2-6x+13\mid pleasecheck$$

Commented by MJS last updated on 02/Jul/19

$$\frac{d}{dx}\left[\ln (x^2-6x+13)\right]=\frac{1}{x^2-6x+13}\times (2x-6)$$$$\mathrm{so}{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{wrong}$$

Commented by Rio Michael last updated on 04/Jul/19

$$yourright.$$

Answered by MJS last updated on 03/Jul/19

$$\int \frac{dx}{4+{(x-3)}^2}=$$$$[t=\frac{x-3}{2}\to dx=2dt]$$$$=\frac{1}{2}\int \frac{dt}{t^2+1}=\frac{1}{2}\arctan t=\frac{1}{2}\arctan \frac{x-3}{2}+C$$

Commented by aliesam last updated on 03/Jul/19

$$godblessyousir$$

Answered by MJS last updated on 03/Jul/19

$$\mathrm{the}{2}^{\mathrm{nd}}\mathrm{one}\mathrm{cannot}\mathrm{be}\mathrm{solved}\mathrm{with}$$$$\tan \left(x^2\right)$$$$\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{easy}\mathrm{like}\mathrm{this}:$$$$\int \frac{{2\tan }^{3}x+2\tan x+1}{\tan x{(1+{\tan }^{2}x)}^2}dx=$$$$[t=\tan x\to dx=dt{\cos }^{2}x]$$$$=\int \frac{2t^3+2t+1}{t{(t^2+1)}^2}dt=\int (-\frac{t}{{(t^2+1)}^2}-\frac{t}{t^2+1}+\frac{2}{t^2+1}+\frac{1}{t})dt=$$$$=\frac{1}{2(t^2+1)}-\frac{1}{2}\ln (t^2+1)+2\arctan t+\ln t$$$$...$$

Commented by aliesam last updated on 03/Jul/19

$$thatsrightitsatypo$$

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