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Question Number 118905 by benjo_mathlover last updated on 20/Oct/20

$$\int \frac{d\lambda }{{({\lambda }^2-9)}^2}=?$$

Answered by Bird last updated on 21/Oct/20

$$I=\int \frac{dx}{{(x^2-9)}^2}\Rightarrow I=\int \frac{dx}{{(x-3)}^2{(x+3)}^2}$$$$=\int \frac{dx}{{\left(\frac{x-3}{x+3}\right)}^2{(x+3)}^4}wedothechangement$$$$\frac{x-3}{x+3}=t\Rightarrow x-3=tx+3t\Rightarrow (1-t)x=3t+3$$$$\Rightarrow x=\frac{3t+3}{1-t}\Rightarrow \frac{dx}{dt}=\frac{3(1-t)+(3t+3)}{{(1-t)}^2}$$$$=\frac{6}{{(1-t)}^2}andx+3=\frac{3t+3}{1-t}+3$$$$=\frac{3t+3+3-3t}{1-t}=\frac{6}{1-t}\Rightarrow$$$$I=\int \frac{6}{{(1-t)}^2{t}^2{\left(\frac{6}{1-t}\right)}^4}dt$$$$=\frac{1}{6^3}\int \frac{{(1-t)}^4}{{(1-t)}^2{t}^2}dt=\frac{1}{6^3}\int \frac{{(1-t)}^2}{t^2}dt$$$$=\frac{1}{6^3}\int \frac{t^2-2t+1}{t^2}dt$$$$=\frac{1}{6^3}\int (1-\frac{2}{t}+\frac{1}{t^2})dt$$$$=\frac{t}{6^3}-\frac{2}{6^3}ln\mid t\mid -\frac{1}{6^{3}t}+C$$$$=\frac{1}{6^3}\{\frac{x-3}{x+3}-2ln\mid \frac{x-3}{x+3}\mid -\frac{x+3}{x-3}\}+C$$

Answered by MJS_new last updated on 20/Oct/20

$$\int \frac{d\lambda }{{({\lambda }^2-9)}^2}=$$$$=\frac{1}{108}\int \frac{1}{x+3}-\frac{1}{x-3}dx+\frac{1}{36}\int \frac{1}{{(x+3)}^2}+\frac{1}{{(x-3)}^2}dx=$$$$=\frac{1}{108}\ln \mid \frac{x+3}{x-3}\mid -\frac{x}{18(x^2-9)}+C$$

Commented by Dwaipayan Shikari last updated on 20/Oct/20

$$Typo$$$$\int \frac{d\lambda }{{({\lambda }^2-9)}^2}$$

Commented by MJS_new last updated on 20/Oct/20

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{the}\mathrm{Master}\mathrm{of}\mathrm{Typos}\mathrm{it}\mathrm{seems}...\mathrm{thank}\mathrm{you}!$$

Commented by Dwaipayan Shikari last updated on 20/Oct/20

$$Youaregettingoldsir!:)Soithappens:)$$

Commented by MJS_new last updated on 20/Oct/20

$$\mathrm{I}\mathrm{got}\mathrm{used}\mathrm{to}\mathrm{getting}\mathrm{older}\mathrm{from}\mathrm{my}\mathrm{first}$$$$\mathrm{breath}...$$

Answered by Dwaipayan Shikari last updated on 20/Oct/20

$$\int \frac{d\lambda }{{({\lambda }^2+9)}^2-36{\lambda }^2}$$$$=\int \frac{1}{12\lambda }(\frac{1}{{\lambda }^2+9-6\lambda }-\frac{1}{{\lambda }^2+9+6\lambda }d\lambda )$$$$=\frac{1}{12}\int \frac{1}{\lambda {(\lambda -3)}^2}-\frac{1}{\lambda {(\lambda +3)}^2}d\lambda$$$$=\frac{1}{12}\int \frac{1}{t^2(t+3)}dt-\frac{1}{u^2(u-3)}du(\lambda -3)=t(\lambda +3)=u$$$$=\frac{1}{36}\int \frac{1}{t}(\frac{1}{t}-\frac{1}{t+3})-\frac{1}{36}\int \frac{1}{u}(\frac{1}{u-3}-\frac{1}{u})$$$$=-\frac{1}{36t}-\frac{1}{108}\int \frac{1}{t}-\frac{1}{t+3}dt-\frac{1}{36u}-\frac{1}{108}\int \frac{1}{u-3}-\frac{1}{u}$$$$=-\frac{1}{36t}+\frac{1}{108}log\left(\frac{t+3}{t}\right)-\frac{1}{36u}+\frac{1}{108}log\left(\frac{u}{u-3}\right)$$$$=-\frac{1}{18}\left(\frac{\lambda }{{\lambda }^2-9}\right)+\frac{1}{108}(log\left(\frac{\lambda }{\lambda -3}\right)+log\left(\frac{\lambda +3}{\lambda }\right))+C$$$$=-\frac{1}{18}\left(\frac{\lambda }{{\lambda }^2-9}\right)+\frac{1}{108}log\left(\frac{\lambda +3}{\lambda -3}\right)+C$$

Answered by bramlexs22 last updated on 20/Oct/20

$$\Rightarrow \frac{1}{{({\lambda }^2-9)}^2}=\frac{a}{\lambda -3}+\frac{b}{{(\lambda -3)}^2}+\frac{c}{(\lambda +3)}+\frac{d}{{(\lambda +3)}^2}$$$$\Rightarrow 1=a(\lambda -3){(\lambda +3)}^2+b{(\lambda +3)}^2+c(\lambda +3){(\lambda -3)}^2+d{(\lambda -3)}^2$$$$\lambda =3\Rightarrow 1=36b;b=\frac{1}{36}$$$$\lambda =-3\Rightarrow 1=36d;d=\frac{1}{36}$$$$\lambda =0\Rightarrow 1=-27a+\frac{1}{4}+27c+\frac{1}{4}$$$$\frac{1}{54}=-a+c...\left(i\right)$$$$\lambda =2\Rightarrow 1=-25a+\frac{25}{36}+5c+\frac{1}{36}$$$$\frac{3}{54}=-5a+c...\left(ii\right)$$$$substract\left(i\right)\mathrm{\&}\left(ii\right)\Rightarrow 4a=-\frac{2}{54};a=-\frac{1}{108}$$$$\Rightarrow c=\frac{1}{54}-\frac{1}{108}=\frac{1}{108}$$$$thenI=-\frac{1}{108}\int \frac{d\lambda }{\lambda -3}+\frac{1}{36}\int \frac{d\lambda }{{(\lambda -3)}^2}+\frac{1}{108}\int \frac{d\lambda }{\lambda +3}+\frac{1}{36}\int \frac{d\lambda }{{(\lambda +3)}^2}$$$$I=\frac{1}{108}\ln \left(\frac{\lambda +3}{\lambda -3}\right)-\frac{1}{36}(\frac{1}{\lambda -3}+\frac{1}{\lambda +3})+c$$$$I=\frac{1}{108}\ln \left(\frac{\lambda +3}{\lambda -3}\right)-\frac{1}{18}\left(\frac{\lambda }{{\lambda }^2-9}\right)+c$$

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