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Question Number 93638 by i jagooll last updated on 14/May/20

$$\int \frac{x^4+4x^2}{\sqrt{x^2+4}}dx?$$

Commented by i jagooll last updated on 14/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{both}$$

Commented by mathmax by abdo last updated on 14/May/20

$$I=\int \frac{x^4+4x^2}{\sqrt{x^2+4}}dxchangement\sqrt{x^2+4}=tgive{x}^2=t^2-4\Rightarrow xdx=tdt$$$$I=\int \frac{{(t^2-4)}^3+4\sqrt{t^2-4}}{t}tdt=\int ({(t^2-4)}^3+4\sqrt{t^2-4})dt$$$$=\int {(t^2-4)}^{3}dt+4\int \sqrt{t^2-4}dtbut$$$$\int {(t^2-4)}^{3}dt=\int (t^6-3t^4\left(4\right)+3t^2{\left(4\right)}^2-4^3)dt$$$$=\frac{t^7}{7}-\frac{12}{5}{t}^5+\frac{48}{3}{t}^3-64t+c_1$$$$\int \sqrt{t^2-4}dt{=}_{t=2chz}\int 2shz\left(2shz\right)dz=4\int {sh}^{2}zdz$$$$=2\int (ch\left(2z\right)-1)dz=sh\left(2z\right)-2z+c_2$$$$=2sh\left(z\right)ch\left(z\right)-2z+c_2$$$$=2\sqrt{{ch}^{2}z-1}\times \frac{t}{2}-2argch\left(\frac{t}{2}\right)+c_2$$$$=t\sqrt{\frac{t^2}{4}-1}-2ln(\frac{t}{2}+\sqrt{\frac{t^2}{4}-1})+c_2\Rightarrow$$$$I=\frac{1}{7}{\left(\sqrt{x^2+4}\right)}^7-\frac{12}{5}{\left(\sqrt{x^2+4}\right)}^5+16{\left(\sqrt{x^2+4}\right)}^3-64\sqrt{x^2+4}$$$$+2t\sqrt{t^2-4}-2ln(t+\sqrt{t^2-4})+C$$$$$$

Answered by Ar Brandon last updated on 14/May/20

$$\mathrm{L}=\int \frac{x^4+4x^2}{\sqrt{x^2+4}}\mathrm{d}x=\int x^2\cdot \frac{x^2+4}{\sqrt{x^2+4}}\mathrm{d}x=\int x^2\sqrt{x^2+4}\mathrm{d}x$$$$x=2\sinh \theta \Rightarrow \mathrm{d}x=2\cosh \theta \mathrm{d}\theta$$$$\mathrm{L}=\int {4\sinh }^2\theta \cdot \sqrt{{4\sinh }^2\theta +4}\cdot 2\cosh \theta \mathrm{d}\theta$$$$\Rightarrow \mathrm{L}=4\int {4\sinh }^2\theta {\cosh }^2\theta \mathrm{d}\theta =4\int {\sinh }^{2}2\theta \mathrm{d}\theta$$$$\Rightarrow \mathrm{L}=2\int (\cosh 4\theta -1)=\frac{\sinh 4\theta }{2}-2\theta +\mathrm{C}$$$$\int \frac{x^4+4x^2}{\sqrt{x^2+4}}dx=\frac{1}{2}{\sinh 4\sinh }^{-1}\left(\frac{x}{2}\right)-{2\sinh }^{-1}\left(\frac{x}{2}\right)+\mathrm{C}$$

Answered by Kunal12588 last updated on 14/May/20

$$I=\int \frac{x^2(x^2+4)}{\sqrt{x^2+4}}dx$$$$x^2+4=t$$$$\Rightarrow dx=\frac{1}{2\sqrt{t-4}}dt$$$$I=\frac{1}{2}\int \frac{(t-4)t}{\sqrt{t}\sqrt{t-4}}dt=\frac{1}{2}\int \sqrt{t}\sqrt{t-4}dt=\frac{1}{2}\int \sqrt{t^2-4t}dt$$$$=\frac{1}{2}\int \sqrt{{(t-2)}^2-4}dt$$$$=\frac{1}{2}[\frac{1}{2}(t-2)\sqrt{t^2-4t}-\frac{4}{2}ln\mid t-2+\sqrt{t^2-4t}\mid ]+C$$$$=\frac{1}{2}[\frac{1}{2}x(x^2+2)\sqrt{x^2+4}-2ln\mid x^2+2+x\sqrt{x^2+4}\mid ]+C$$$$=\frac{1}{4}x(x^2+2)\sqrt{x^2+4}-ln\mid x^2+x\sqrt{x^2+4}+2\mid +C$$

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