The Notes: `int_0^1 sqrt(x^2 + 1) dx` Evaluate the integral

$\displaystyle{\int_{{0}}^{{1}}}\sqrt{{{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}$

Let's first evaluate the definite integral using the standard integral,

$\displaystyle\int\sqrt{{{{a}}^{{2}}+{{x}}^{{2}}}}{\left.{d}{x}\right.}=\frac{{{x}\sqrt{{{{a}}^{{2}}+{{x}}^{{2}}}}}}{{2}}+\frac{{{a}}^{{2}}}{{2}}{\ln}{\left|{x}+\sqrt{{{{a}}^{{2}}+{{x}}^{{2}}}}\right|}+{C}$

$\displaystyle\int\sqrt{{{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}=\frac{{{x}\sqrt{{{{1}}^{{2}}+{{x}}^{{2}}}}}}{{2}}+\frac{{{1}}^{{2}}}{{2}}{\ln}{\left|{x}+\sqrt{{{{1}}^{{2}}+{{x}}^{{2}}}}\right|}+{C}$

$\displaystyle=\frac{{{x}\sqrt{{{1}+{{x}}^{{2}}}}}}{{2}}+\frac{{1}}{{2}}{\ln}{\left|{x}+\sqrt{{{1}+{{x}}^{{2}}}}\right|}+{C}$

Now let's evaluate the definite integral,

$\displaystyle{\int_{{0}}^{{1}}}\sqrt{{{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}={{\left[\frac{{{x}\sqrt{{{1}+{{x}}^{{2}}}}}}{{2}}+\frac{{1}}{{2}}{\ln}{\left|{x}+\sqrt{{{1}+{{x}}^{{2}}}}\right|}\right]}_{{0}}^{{1}}}$

$\displaystyle={\left[\frac{{{1}\sqrt{{{1}+{1}}}}}{{2}}+\frac{{1}}{{2}}{\ln}{\left|{1}+\sqrt{{{1}+{1}}}\right|}\right]}-{\left[\frac{{{0}\sqrt{{{1}+{0}}}}}{{2}}+\frac{{1}}{{2}}{\ln}{\left|{0}+\sqrt{{{1}+{0}}}\right|}\right]}$

$\displaystyle={\left[\frac{\sqrt{{{2}}}}{{2}}+\frac{{1}}{{2}}{\ln}{\left|{1}+\sqrt{{{2}}}\right|}\right]}-{\left[{0}+\frac{{1}}{{2}}{\ln{{\left({1}\right)}}}\right]}$

$\displaystyle=\frac{\sqrt{{{2}}}}{{2}}+\frac{{1}}{{2}}{\ln{{\left({1}+\sqrt{{{2}}}\right)}}}$

Since $\displaystyle{\arcsin{{h}}}{\left({x}\right)}={\ln{{\left({x}+\sqrt{{{{x}}^{{2}}+{1}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{\left(\sqrt{{{2}}}+{\arcsin{{h}}}{\left({1}\right)}\right)}$

The Notes

Thursday, May 9, 2013

$\displaystyle{\int_{{0}}^{{1}}}\sqrt{{{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}$ Evaluate the integral

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