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Question Number 83675 by niroj last updated on 05/Mar/20

$$$$$$$$$$\mathrm{evaluate}:$$$$2{\int }_0^2\frac{\sqrt{\mathrm{x}+1}}{{\mathrm{x}}^2+4}\mathrm{dx}$$$$$$$$$$

Commented by jagoll last updated on 05/Mar/20

$$\sqrt{x+1}=u\Rightarrow x^2={(u^2-1)}^2$$$$2xdx=4u(u^2-1)du$$$$dx=\frac{2u(u^2-1)du}{u^2-1}=2udu$$$$x^2+4={(u^2-1)}^2+4$$$$2{\stackrel{\sqrt{3}}{\int }}_1\frac{u\left(2udu\right)}{4+{(u^2-1)}^2}$$$$letu=\sec t\Rightarrow \mathrm{du}=\sec t\tan tdt$$$$$$

Commented by niroj last updated on 05/Mar/20

$$thanksmr.jagolltrytosolveifitisbettertobecomplete.$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

Answered by TANMAY PANACEA last updated on 05/Mar/20

$$I=\int \frac{x+1}{\sqrt{x+1}\times (x^2+4)}dx$$$$t^2=x+1$$$$\int \frac{t^2\times 2tdt}{t(t^4-2t^2+1+4)}$$$$2\int \frac{t^2}{t^4-2t^2+5}dt$$$$2\int \frac{dt}{t^2+\frac{5}{t^2}-2}$$$$t^2+\frac{5}{t^2}={(t+\frac{\sqrt{5}}{t})}^2-2\sqrt{5}$$$$\frac{d}{dt}(t+\frac{\sqrt{5}}{t})=1-\frac{\sqrt{5}}{t^2}$$$$t^2+\frac{5}{t^2}={(t-\frac{\sqrt{5}}{t})}^2+2\sqrt{5}$$$$\frac{d}{dt}(t-\frac{\sqrt{5}}{t})=1+\frac{\sqrt{5}}{t^2}$$$$look(1+\frac{\sqrt{5}}{t^2})+(1-\frac{\sqrt{5}}{t^2})=2$$$$now\int \frac{1-\frac{\sqrt{5}}{t^2}+1+\frac{\sqrt{5}}{t^2}}{t^2+\frac{5}{t^2}-2}dt$$$$\int \frac{d(t+\frac{\sqrt{5}}{t})}{{(t+\frac{\sqrt{5}}{t})}^2-2\sqrt{5}-2}+\int \frac{d(t-\frac{\sqrt{5}}{t})}{{(t-\frac{\sqrt{5}}{t})}^2+2\sqrt{5}-2}$$$$\mathit{n}\mathit{o}\mathit{w}\mathit{p}\mathit{l}\mathit{s}\mathit{u}\mathit{s}\mathit{e}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}\mathit{o}\mathit{f}$$$$\int \frac{dx}{x^2-a^2}and\int \frac{dx}{x^2+a^2}$$$$2\sqrt{5}+2={\left(\sqrt{2\sqrt{5}+2}\right)}^2\leftarrow asif{a}^2$$$$2\sqrt{5}-2={\left(\sqrt{2\sqrt{5}-2}\right)}^2$$$$so$$$$\frac{1}{2\left(\sqrt{2\sqrt{5}+2}\right)}ln\left(\frac{(t+\frac{\sqrt{5}}{t})-\left(\sqrt{2\sqrt{5}+2}\right)}{(t+\frac{\sqrt{5}}{t})+\left(\sqrt{2\sqrt{5}+2}\right)}\right)+\frac{1}{\sqrt{2\sqrt{5}-2}}{tan}^{-1}\left(\frac{t-\frac{\sqrt{5}}{t}}{\sqrt{2\sqrt{5}-2}}\right)$$$$\mathit{n}\mathit{o}\mathit{w}\mathit{p}\mathit{l}\mathit{s}\mathit{r}\mathit{e}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}\mathit{t}\mathit{b}\mathit{y}\sqrt{\mathit{x}+1}\mathit{a}\mathit{n}\mathit{d}\mathit{p}\mathit{u}\mathit{t}\mathit{u}\mathit{p}\mathit{p}\mathit{e}\mathit{r}$$$$\mathit{a}\mathit{n}\mathit{d}\mathit{l}\mathit{o}\mathit{w}\mathit{e}\mathit{r}\mathit{l}\mathit{i}\mathit{m}\mathit{i}\mathit{t}$$$$$$

Commented by niroj last updated on 05/Mar/20

$$youareabsuteamazing.nicesolution..$$

Commented by TANMAY PANACEA last updated on 05/Mar/20

$$thankyousir$$

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