`int x/sqrt(x^2 + x + 1) dx` Evaluate the integral

Sunday, January 7, 2018

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

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

Let's rewrite the integrand by completing the square of the denominator,

$\displaystyle=\int\frac{{x}}{\sqrt{{{{\left({x}+\frac{{1}}{{2}}\right)}}^{{2}}+\frac{{3}}{{4}}}}}{\left.{d}{x}\right.}$

Now let's apply the integral
substitution,

Let $\displaystyle{u}={x}+\frac{{1}}{{2}}$

$\displaystyle{x}={u}-\frac{{1}}{{2}}$

du=1dx

$\displaystyle=\int\frac{{{u}-\frac{{1}}{{2}}}}{\sqrt{{{{u}}^{{2}}+\frac{{3}}{{4}}}}}{d}{u}$

$\displaystyle=\int\frac{{{2}{u}-{1}}}{\sqrt{{{4}{{u}}^{{2}}+{3}}}}{d}{u}$

Now apply the sum rule,

$\displaystyle=\int\frac{{{2}{u}}}{\sqrt{{{4}{{u}}^{{2}}+{3}}}}{d}{u}-\int\frac{{1}}{\sqrt{{{4}{{u}}^{{2}}+{3}}}}{d}{u}$

$\displaystyle={2}\int\frac{{u}}{\sqrt{{{4}{{u}}^{{2}}+{3}}}}{d}{u}-\int\frac{{1}}{\sqrt{{{4}{{u}}^{{2}}+{3}}}}{d}{u}$

Now let's evaluate the
first integral by applying the integral substitution,

Let $\displaystyle{v}={4}{{u}}^{{2}}+{3}$

$\displaystyle{d}{v}={8}{u}{d}{u}$

$\displaystyle\int\frac{{u}}{\sqrt{{{4}{{u}}^{{2}}+{3}}}}{d}{u}=\int\frac{{1}}{{{8}\sqrt{{{v}}}}}{d}{v}$

$\displaystyle=\frac{{1}}{{8}}\int{{v}}^{{-\frac{{1}}{{2}}}}{d}{v}$

$\displaystyle=\frac{{1}}{{8}}\frac{{{{v}}^{{-\frac{{1}}{{2}}+{1}}}}}{{-\frac{{1}}{{2}}+{1}}}$

$\displaystyle=\frac{{1}}{{8}}\frac{{{v}}^{{\frac{{1}}{{2}}}}}{{\frac{{1}}{{2}}}}$

$\displaystyle=\frac{{2}}{{8}}{{v}}^{{\frac{{1}}{{2}}}}$

$\displaystyle=\frac{{1}}{{4}}\sqrt{{{v}}}$

substitute back $\displaystyle{v}={4}{{u}}^{{2}}+{3}$

$\displaystyle=\frac{{1}}{{4}}\sqrt{{{4}{{u}}^{{2}}+{3}}}$

Now let's evaluate the second integral
$\displaystyle\int\frac{{1}}{\sqrt{{{4}{{u}}^{{2}}+{3}}}}{d}{u}$ using integral substitution,

For $\displaystyle\sqrt{{{b}{{x}}^{{2}}+{a}}}$
substitute $\displaystyle{x}=\frac{\sqrt{{{a}}}}{\sqrt{{{b}}}}{\tan{{\left({v}\right)}}}$ ,

Let $\displaystyle{u}=\frac{\sqrt{{{3}}}}{{2}}{\tan{{\left({v}\right)}}}$

$\displaystyle{d}{u}=\frac{\sqrt{{{3}}}}{{2}}{{\sec}}^{{2}}{\left({v}\right)}{d}{v}$

$\displaystyle\int\frac{{1}}{\sqrt{{{4}{{v}}^{{2}}+{3}}}}{d}{u}=\int\frac{{\frac{\sqrt{{{3}}}}{{2}}{{\sec}}^{{2}}{\left({v}\right)}}}{\sqrt{{{4}{{\left(\frac{\sqrt{{{3}}}}{{2}}{\tan{{\left({v}\right)}}}\right)}}^{{2}}+{3}}}}{d}{v}$

$\displaystyle=\int\frac{{\sqrt{{{3}}}{{\sec}}^{{2}}{\left({v}\right)}}}{{{2}\sqrt{{{3}{{\tan}}^{{2}}{\left({v}\right)}+{3}}}}}{d}{v}$

$\displaystyle=\frac{\sqrt{{{3}}}}{{2}}\int\frac{{{{\sec}}^{{2}}{\left({v}\right)}}}{\sqrt{{{3}{{\tan}}^{{2}}{\left({v}\right)}+{3}}}}{d}{v}$

$\displaystyle=\frac{\sqrt{{{3}}}}{{2}}\int\frac{{{{\sec}}^{{2}}{\left({v}\right)}}}{{\sqrt{{{3}}}\sqrt{{{{\tan}}^{{2}}+{1}}}}}{d}{v}$

$\displaystyle=\frac{{1}}{{2}}\int\frac{{{{\sec}}^{{2}}{\left({v}\right)}}}{\sqrt{{{{\tan}}^{{2}}{\left({v}\right)}+{1}}}}{d}{v}$

Now use the
identity: $\displaystyle{1}+{{\tan}}^{{2}}{\left({x}\right)}={{\sec}}^{{2}}{\left({x}\right)}$

$\displaystyle=\frac{{1}}{{2}}\int\frac{{{{\sec}}^{{2}}{\left({v}\right)}}}{\sqrt{{{{\sec}}^{{2}}{\left({v}\right)}}}}{d}{v}$

assuming sec(v) $\displaystyle\ge{0}$

$\displaystyle=\frac{{1}}{{2}}\int{\sec{{\left({v}\right)}}}{d}{v}$

Now using the common integral,

$\displaystyle\int{\sec{{\left({v}\right)}}}{\left.{d}{x}\right.}={\ln{{\left({\left({\sec{{\left({v}\right)}}}+{\tan{{\left({v}\right)}}}\right)}\right.}}}$

$\displaystyle=\frac{{1}}{{2}}{\left({\ln{{\left({\sec{{\left({v}\right)}}}+{\tan{{\left({v}\right)}}}\right)}}}\right.}$

Substitute back $\displaystyle{v}={\arctan{{\left(\frac{{{2}{u}}}{\sqrt{{{3}}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{\left[{\ln{{\left\lbrace{\sec{{\left({\arctan{{\left(\frac{{{2}{u}}}{\sqrt{{{3}}}}\right)}}}\right)}}}+{\tan{{\left({\arctan{{\left(\frac{{{2}{u}}}{\sqrt{{{3}}}}\right)}}}\right\rbrace}}}\right]}}}\right.}$

$=1/2[ln{sqrt(1+(4u^2)/3)+(2u)/sqrt(3)}]$

$\displaystyle\int\frac{{{2}{u}-{1}}}{\sqrt{{{4}{{u}}^{{2}}+{3}}}}{d}{u}={2}{\left(\frac{{1}}{{4}}\sqrt{{{4}{{u}}^{{2}}+{3}}}\right)}-\frac{{1}}{{2}}{\ln{{\left(\sqrt{{{1}+{4}\frac{{{u}}^{{2}}}{{3}}}}+\frac{{{2}{u}}}{\sqrt{{{3}}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}\sqrt{{{4}{{u}}^{{2}}+{3}}}-\frac{{1}}{{2}}{\ln{{\left(\sqrt{{{1}+\frac{{{4}{{u}}^{{2}}}}{{3}}}}+\frac{{{2}{u}}}{\sqrt{{{3}}}}\right)}}}$

Substitute
back $\displaystyle{u}={x}+\frac{{1}}{{2}}$

$\displaystyle=\frac{{1}}{{2}}\sqrt{{{4}{{\left({x}+\frac{{1}}{{2}}\right)}}^{{2}}+{3}}}-\frac{{1}}{{2}}{\ln{{\left(\sqrt{{{1}+\frac{{{4}{{\left({x}+\frac{{1}}{{2}}\right)}}^{{2}}}}{{3}}}}+\frac{{{2}{\left({x}+\frac{{1}}{{2}}\right)}}}{\sqrt{{{3}}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}\sqrt{{{4}{\left({{x}}^{{2}}+\frac{{1}}{{4}}+{x}\right)}+{3}}}-\frac{{1}}{{2}}{\ln{{\left(\sqrt{{{1}+\frac{{4}}{{3}}{\left({{x}}^{{2}}+\frac{{1}}{{4}}+{x}\right)}}}+{\left(\frac{{2}}{\sqrt{{{3}}}}\right)}\frac{{{2}{x}+{1}}}{{2}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}\sqrt{{{4}{{x}}^{{2}}+{1}+{4}{x}+{3}}}-\frac{{1}}{{2}}{\ln{{\left(\sqrt{{\frac{{{3}+{4}{{x}}^{{2}}+{1}+{4}{x}}}{{3}}}}+\frac{{{2}{x}+{1}}}{\sqrt{{{3}}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}\sqrt{{{4}{{x}}^{{2}}+{4}{x}+{4}}}-\frac{{1}}{{2}}{\ln{{\left(\sqrt{{\frac{{{4}{{x}}^{{2}}+{4}{x}+{4}}}{{3}}}}+\frac{{{2}{x}+{1}}}{\sqrt{{{3}}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}\sqrt{{{4}{\left({{x}}^{{2}}+{x}+{1}\right)}}}-\frac{{1}}{{2}}{\ln{{\left({\left(\frac{{2}}{\sqrt{{{3}}}}\right)}\sqrt{{{{x}}^{{2}}+{x}+{1}}}+\frac{{{2}{x}+{1}}}{\sqrt{{{3}}}}\right)}}}$

$\displaystyle=\sqrt{{{{x}}^{{2}}+{x}+{1}}}-\frac{{1}}{{2}}{\ln{{\left(\frac{{{2}\sqrt{{{{x}}^{{2}}+{x}+{1}}}+{2}{x}+{1}}}{\sqrt{{{3}}}}\right)}}}$

add a constant
C to the solution,

$\displaystyle=\sqrt{{{{x}}^{{2}}+{x}+{1}}}-\frac{{1}}{{2}}{\ln{{\left(\frac{{{2}\sqrt{{{{x}}^{{2}}+{x}+{1}}}+{2}{x}+{1}}}{\sqrt{{{3}}}}\right)}}}+{C}$

Spooky Notes

Sunday, January 7, 2018

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

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