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Question Number 73482 by abdomathmax last updated on 13/Nov/19

$$find\int \frac{dx}{\sqrt{x^2+1}+\sqrt{x^2+3}}$$

Commented by abdomathmax last updated on 17/Nov/19

$$letI=\int \frac{dx}{\sqrt{x^2+1}+\sqrt{x^2+3}}\Rightarrow$$$$I=\int \frac{\sqrt{x^2+3}-\sqrt{x^2+1}}{2}dx=\frac{1}{2}\int \sqrt{x^2+3}dx-\frac{1}{2}\int \sqrt{x^2+1}dx$$$$wehave\int \sqrt{x^2+3}dx{=}_{x=\sqrt{3}sh\left(t\right)}\sqrt{3}\int \sqrt{3}ch\left(t\right)ch\left(t\right)$$$$=3\int \frac{1+ch\left(2t\right)}{2}dt=\frac{3}{2}t+\frac{3}{4}sh\left(2t\right)+c_1$$$$=\frac{3}{4}t+\frac{3}{4}2sh\left(t\right)ch\left(t\right)+c_1$$$$=\frac{3}{4}argsh\left(\frac{x}{\sqrt{3}}\right)+\frac{3}{2}\frac{x}{\sqrt{3}}\sqrt{1+\frac{x^2}{3}}+c$$$$=\frac{3}{4}ln(\frac{x}{\sqrt{3}}+\sqrt{1+\frac{x^2}{3}})+\frac{x}{2}\sqrt{3+x^2}+c_1$$$$\int \sqrt{x^2+1}dx{=}_{x=sh\left(t\right)}\int ch\left(t\right)ch\left(t\right)dt$$$$=\int \frac{1+ch\left(2t\right)}{2}dt=\frac{t}{2}+\frac{1}{4}sh\left(2t\right)+c_2$$$$=\frac{t}{2}+\frac{1}{2}sh\left(g\right)ch\left(t\right)+c_2$$$$=\frac{1}{2}ln(x+\sqrt{1+x^2})+\frac{x}{2}\sqrt{1+x^2}+c_2\Rightarrow$$$$I=\frac{3}{8}ln(x+\sqrt{3+x^2})+\frac{x}{4}\sqrt{3+x^2}-\frac{1}{4}ln(x+\sqrt{1+x^2})$$$$-\frac{x}{4}\sqrt{1+x^2}+C$$$$$$

Answered by ajfour last updated on 13/Nov/19

$$I=\int \frac{dx}{\sqrt{x^2+1}+\sqrt{x^2+3}}$$$$=\frac{1}{2}\int (\sqrt{x^2+3}-\sqrt{x^2+1})dx$$$$=\frac{x}{4}(\sqrt{x^2+3}-\sqrt{x^2+1})$$$$+\frac{3}{4}\ln \mid x+\sqrt{x^2+3}\mid -\frac{1}{4}\ln \mid x+\sqrt{x^2+1}\mid +c$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}.$$

Commented by abdomathmax last updated on 17/Nov/19

$$thankssir.$$

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