New Notes Blog: `int 1/((x-1)sqrt(4x^2-8x+3)) dx` Find the indefinite integral

Friday, 22 August 2014

$\displaystyle\int\frac{{1}}{{{\left({x}-{1}\right)}\sqrt{{{4}{{x}}^{{2}}-{8}{x}+{3}}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

$\displaystyle\int\frac{{1}}{{{\left({x}-{1}\right)}\sqrt{{{4}{{x}}^{{2}}-{8}{x}+{3}}}}}{\left.{d}{x}\right.}$

Let's rewrite the integral by completing the square of the term in denominator,

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

Apply the integral substitution: $\displaystyle{u}={2}{x}-{2}$

$\displaystyle{d}{u}={2}{\left.{d}{x}\right.}$

$\displaystyle\Rightarrow{\left.{d}{x}\right.}=\frac{{{d}{u}}}{{2}}$

$\displaystyle{u}={2}{\left({x}-{1}\right)}$

$\displaystyle\Rightarrow{\left({x}-{1}\right)}=\frac{{u}}{{2}}$

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

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

Again apply integral substitution: $\displaystyle{u}={\sec{{\left({v}\right)}}}$

$\displaystyle{d}{u}={\sec{{\left({v}\right)}}}{\tan{{\left({v}\right)}}}{d}{v}$

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

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

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

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

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

$\displaystyle=\int\frac{{\tan{{\left({v}\right)}}}}{{\tan{{\left({v}\right)}}}}{d}{v}$ assuming $\displaystyle{\tan{{\left({v}\right)}}}\ge{0}$

$\displaystyle=\int{d}{v}$

$\displaystyle={v}$

Substitute back $\displaystyle{v}={a}{r}{c}{\sec{{\left({u}\right)}}}$ and $\displaystyle{u}={\left({2}{x}-{2}\right)}$

and add a constant C to the solution,

$\displaystyle={a}{r}{c}{\sec{{\left({2}{x}-{2}\right)}}}+{C}$