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Question Number 123452 by Eric002 last updated on 25/Nov/20

$$\int \frac{dx}{(x^2+n)\sqrt{x^2+a}}$$

Commented by MJS_new last updated on 25/Nov/20

$$\mathrm{depends}\mathrm{on}\mathrm{the}\mathrm{values}\mathrm{of}a\mathrm{and}n\mathrm{and}\mathrm{their}$$$$\mathrm{relation}...$$

Answered by TANMAY PANACEA last updated on 25/Nov/20

$$t^2=\frac{x^2+a}{x^2+n}$$$$t^2{x}^2+t^{2}n=x^2+a$$$$x^2(t^2-1)=a-t^{2}n$$$$x^2=\frac{a-t^{2}n}{t^2-1}$$$$2xdx=\frac{(t^2-1)(-2tn)-(a-t^{2}n)\left(2t\right)}{{(t^2-1)}^2}dt$$$$2xdx=\frac{-2t^{3}n+2tn-2at+2t^{3}n}{{(t^2-1)}^2}dt=\frac{2t(n-a)}{{(t^2-1)}^2}dt$$$$xdx=\frac{t(n-a)}{{(t^2-1)}^2}dt$$$$dx=\frac{t(n-a)}{{(t^2-1)}^2}\times \frac{1}{x}=\frac{t(n-a)}{{(t^2-1)}^2}\times \frac{\sqrt{t^2-1}}{\sqrt{a-t^{2}n}}dt$$$$x^2+n$$$$=\frac{a-t^{2}n}{t^2-1}+n=\frac{a-t^{2}n+{nt}^2-n}{t^2-1}=\frac{a-n}{t^2-1}$$$$x^2+a=\frac{a-t^{2}n}{t^2-1}+a=\frac{a-t^{2}n+{at}^2-a}{t^2-1}=\frac{t^2(a-n)}{t^2-1}$$$$\int \frac{dx}{(x^2+n)\sqrt{x^2+a}}$$$$\int \frac{t(a-n)}{{(t^2-1)}^2}\times \sqrt{t^2-1}\times \frac{1}{\sqrt{a-t^{2}n}}\times \frac{t^2-1}{a-n}\times \frac{1}{\sqrt{\frac{t^2(a-n)}{t^2-1}}}\times dt$$$$\int \frac{t(a-n)}{{(t^2-1)}^2}\times {(t^2-1)}^2\times \frac{1}{\sqrt{a-t^{2}n}}\times \frac{1}{{(a-n)}^{\frac{3}{2}}}\times \frac{dt}{t}$$$$\int \frac{1}{{(a-n)}^{\frac{1}{2}}}\times \frac{dt}{\sqrt{n}\left(\sqrt{\frac{a}{n}-t^2}\right)}$$$$\frac{1}{{(an-n^2)}^{\frac{1}{2}}}\int \frac{dt}{\sqrt{\frac{a}{n}-t^2}}$$$$$$$$\frac{1}{{(an-n^2)}^{\frac{1}{2}}}{sin}^{-1}\left(\frac{t}{\sqrt{\frac{a}{n}}}\right)+C$$$$\frac{1}{{(an-n^2)}^{\frac{1}{2}}}\times {sin}^{-1}\left(\frac{\sqrt{\frac{x^2+a}{x^2+n}}}{\sqrt{\frac{a}{n}}}\right)+C$$$$plschk$$

Commented by TANMAY PANACEA last updated on 25/Nov/20

$$pldchkerrorifsny$$

Commented by MJS_new last updated on 25/Nov/20

$$\mathrm{what}\mathrm{if}an-n^2⩽0\mathrm{or}\frac{a}{n}<0?$$

Answered by nueron last updated on 25/Nov/20

Commented by MJS_new last updated on 25/Nov/20

$$\mathrm{yeah}.\mathrm{but}\mathrm{for}a=-4\wedge n=3\mathrm{you}\mathrm{would}$$$$\mathrm{probably}\mathrm{choose}\mathrm{a}\mathrm{different}\mathrm{path}...$$

Answered by TANMAY PANACEA last updated on 25/Nov/20

$$x=\sqrt{a}tan\theta$$$$dx=\sqrt{a}{sec}^2\theta d\theta$$$$\int \frac{\sqrt{a}{sec}^2\theta d\theta }{({atan}^2\theta +n)\sqrt{a}sec\theta }$$$$\int \frac{d\theta }{cos\theta (a\frac{{sin}^2\theta }{{cos}^2\theta }+n)}$$$$\int \frac{cos\theta d\theta }{{asin}^2\theta +n(1-{sin}^2\theta )}$$$$\int \frac{d\left(sin\theta \right)}{n+(a-n){sin}^2\theta }$$$$\frac{1}{(a-n)}\int \frac{d\left(sin\theta \right)}{{\left(\sqrt{\frac{n}{a-n}}\right)}^2+{sin}^2\theta }$$$$\frac{1}{(a-n)}\times \frac{\sqrt{a-n}}{\sqrt{n}}{tan}^{-1}\left(\frac{sin\theta }{\sqrt{\frac{n}{a-n}}}\right)+C$$$$\frac{1}{\sqrt{n}\times \sqrt{a-n}}\times {tan}^{-1}\left(\frac{\frac{x}{\sqrt{a+x^2}}}{\sqrt{\frac{n}{a-n}}}\right)+C$$$$$$

Answered by Bird last updated on 25/Nov/20

$$I=\int \frac{dx}{(x^2+n)\sqrt{x^2+a}}$$$$a>0wedothechangement$$$$x=\sqrt{a}sh\left(t\right)\Rightarrow$$$$I=\int \frac{\sqrt{a}ch\left(t\right)}{\sqrt{a}ch\left(t\right)({ash}^{2}t+n)}dt$$$$=\int \frac{dt}{{ash}^{2}t+n}=\int \frac{dt}{a.\frac{ch\left(2t\right)-1}{2}+n}$$$$=\int \frac{2dt}{ach\left(2t\right)-a+2n}$$$$=\int \frac{2dt}{a\frac{e^{2t}+e^{-2t}}{2}+2n-a}$$$$=\int \frac{4dt}{a(e^{2t}+e^{-2t})+4n-2a}$$$${=}_{e^{2t}=u}4\int \frac{du}{2u\{au+{au}^{-1}+4n-2a\}}$$$$=2\int \frac{du}{{au}^2+a+(4n-2a)u}$$$$=2\int \frac{du}{{au}^2+(4n-2a)u+a}$$$${\mathrm{\Delta }}^{{}^{\prime }}={(2n-a)}^2-a^2$$$$=4n^2-4na=4n(n-a)$$$$ifn<a\Rightarrow {\mathrm{\Delta }}^{{}^{\prime }}<0\Rightarrow$$$${au}^2+(4n-2a)u+a$$$$=a\{u^2+2\left(\frac{2n-a)}{a}\right)u+\frac{{(2n-a)}^2}{a^2}$$$$+a-\frac{{(2n-a)}^2}{a^2}\}$$$$=a\{{(u+\frac{2n-a}{a})}^2+\frac{a^3-{(2n-a)}^2}{a^2}\}$$$$inthiscasewedothechangement$$$$u+\frac{2n-a}{a}=\sqrt{\frac{a^3-{(2n-a)}^2}{a^2}}z)$$$$ifn>a{\mathrm{\Delta }}^{{}^{\prime }}>0\Rightarrow u_1=\frac{a-2n+2\sqrt{n^2-na}}{a}$$$$u_2=\frac{a-2n-2\sqrt{n^2-na}}{a}$$$$\Rightarrow I=\frac{2}{a}\int \frac{du}{(u-u_1)(u-u_2)}$$$$=\frac{2}{a(u_1-u_2)}\int (\frac{1}{u-u_1}-\frac{1}{u-u_2})du$$$$=\frac{2}{a(u_1-u_2)}ln\mid \frac{u-u_1}{u-u_2}\mid +c$$$$u=e^{2t}sndt=argsh\left(\frac{x}{\sqrt{a}}\right)$$$$=ln(\frac{x}{\sqrt{a}}+\sqrt{1+\frac{x^2}{a}})$$$$=ln\left(\frac{x+\sqrt{a+x^2}}{\sqrt{a}}\right)\Rightarrow$$$$u={\left(\frac{x+\sqrt{a+x^2}}{\sqrt{a}}\right)}^2....$$

Commented by Bird last updated on 25/Nov/20

$$ifa<0wedothechangement$$$$x=\sqrt{-a}ch\left(t\right)$$

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