Integriranje iracionalnih funkcija racionalnom supstitucijom

Ovakav integral rješavamo supstitucijom $x=t^{k}$ , gdje je

najmanji zajednički višekratnik nazivnika eksponenata od

koji se pojavljuje u podintergalnoj funkciji:

$\displaystyle \int \frac{ dx}{x\left( 1+2\sqrt{x}+\sqrt[3]{x}\right) }$	$\displaystyle ==\left\{ \begin{array}{c} x=t^{6} dx=6t^{5} dt \end{array} \right\}$
	$\displaystyle =\int \frac{6t^{5} dt}{t^{6}\left( 1+2t^{3}+t^{2}\right) }=\int \frac{6 dt}{ t\left( 1+2t^{3}+t^{2}\right) }$
	$\displaystyle =\left\{ \begin{array}{c} \left( 2t^{3}+t^{2}+1\right) =0\Longrig... ...1\right) :\left( t+1\right) =2t^{2}-t+1 \vdots ost.0 \end{array} \right\}$
	$\displaystyle =6\int \frac{ dt}{t\left( t+1\right) \left( 2t^{2}-t+1\right) }$
	$\displaystyle =\left\{ \begin{array}{c} \frac{1}{t\left( t+1\right) \left( 2t^{... ...=1, B=\frac{-1}{4}, C=\frac{-3}{2}, D=\frac{1}{4} \end{array} \right\}$
	$\displaystyle =6\int \frac{ dt}{t}+6\int \frac{\frac{-1}{4} dt}{t+1}-9\int \frac{t-\frac{1 }{6}}{2t^{2}-t+1} dt$
	$\displaystyle =6\ln \left\vert t\right\vert -\frac{3}{2\ln }\left\vert t+1\right\vert -I_{1}$
	$\displaystyle =6\ln \left\vert \sqrt[6]{x}\right\vert -\frac{3}{2\ln }\left\vert \sqrt[6] {x}+1\right\vert -I_{1}$

Integral označen sa $I_{1}$ računamo posebno kao integral racionalne funkcije:

$\displaystyle I_{1}$	$\displaystyle =\int \frac{t-\frac{1}{6}}{2t^{2}-t+1} dt=\frac{1}{4}\int \frac{4t- \frac{2}{3}}{2t^{2}-t+1} dt$
	$\displaystyle =\frac{1}{4}\int \frac{4t-1}{2t^{2}-t+1} dt+\frac{1}{4}\int \frac{\frac{1}{3} }{2t^{2}-t+1} dt$
	$\displaystyle =\frac{1}{4}\ln \left( 2t^{2}-t+1\right) +\frac{1}{24}\int \frac{1}{t^{2}- \frac{1}{2}t+\frac{1}{2}} dt$
	$\displaystyle =\frac{1}{4}\ln \left( 2t^{2}-t+1\right) +\frac{1}{24}\int \frac{1}{\left( t-\frac{1}{4}\right) ^{2}+\frac{7}{16}} dt$
	$\displaystyle =\frac{1}{4}\ln \left( 2t^{2}-t+1\right) +\frac{1}{24}\frac{4}{\sqrt{7}} arctg\frac{4t-1}{\sqrt{7}}+C$
	$\displaystyle =\frac{1}{4}\ln \left( 2t^{2}-t+1\right) +\frac{1}{6\sqrt{7}}arctg\frac{ 4t-1}{\sqrt{7}}+C$
	$\displaystyle =\frac{1}{4}\ln \left( 2\sqrt[3]{x}-\sqrt[6]{x}+1\right) +\frac{1}{6\sqrt{7 }}arctg\frac{4\sqrt[6]{x}-1}{\sqrt{7}}+C$

pa je konačno rješenje

$\displaystyle \int \frac{ dx}{x\left( 1+2\sqrt{x}+\sqrt[3]{x}\right) }$	$\displaystyle =6\ln \left\vert \sqrt[6]{x}\right\vert -\frac{3}{2\ln }\left\ver... ...rt[6]{x}+1\right\vert - \frac{1}{4}\ln \left( 2\sqrt[3]{x}-\sqrt[6]{x}+1\right)$
	$\displaystyle \quad -\frac{1}{6\sqrt{7}} \mathop{\mathrm{arctg}}\nolimits \frac{4\sqrt[6]{x}-1}{\sqrt{7}}+C.$

Vrijedi

$\displaystyle \int \frac{ dx}{\left( 2x+1\right) ^{\frac{2}{3}}-\left( 2x+1\right) ^{\frac{1 }{2}}}$	$\displaystyle =\left\{ \begin{array}{cc} 2x+1=t^{6} & dx=3t^{5} x=\frac{t^{6}-1}{2} & t=\sqrt[6]{2x+1} \end{array} \right\}$
	$\displaystyle =\int \frac{3t^{5} dt}{t^{4}-t^{3}}=3\int \frac{t^{2} dt}{t-1}=3\int \frac{ t^{2}-1+1}{t-1} dt$
	$\displaystyle =3\int \left( t+1+\frac{1}{t-1}\right) dt$
	$\displaystyle =3\int \left( t+1\right) dt+3\int \frac{1}{t-1} dt$
	$\displaystyle =\frac{3}{2}t^{2}+3t+3\ln \left\vert t-1\right\vert +C$
	$\displaystyle =\frac{3}{2}\sqrt[3]{2x+1}+3\sqrt[6]{2x+1}+3\ln \left\vert \sqrt[6]{2x+1} -1\right\vert +C.$

Vrijedi

$\displaystyle \int \frac{ dx}{\sqrt[4]{\left( x-1\right) ^{3}\left( x+2\right) ^{5}}}$	$\displaystyle =\int \frac{ dx}{\sqrt[4]{\left( \frac{x+2}{x-1}\right) \left( x-1\right) ^{4}\left( x+2\right) ^{4}}}$
	$\displaystyle =\int \frac{1}{\left( x-1\right) \left( x+2\right) }\sqrt[4]{\left( \frac{ x-1}{x+2}\right) } dx$
	$\displaystyle =\left\{ \begin{array}{cc} \frac{x-1}{x+2}=t^{4} & dx=\frac{12t... ...( 1-t^{4}\right) ^{2}} dt x=\frac{1+2t^{4}}{1-t^{4}} & \end{array} \right\}$
	$\displaystyle =\int \frac{1-t^{4}}{3}\cdot \frac{1-t^{4}}{3t^{4}}t\frac{12t^{3}}{\left( 1-t^{4}\right) ^{2}} dt$
	$\displaystyle =\int \frac{4}{3} dt=\frac{4}{3}t+C=\frac{4}{3}\sqrt[4]{\frac{x-1}{x+2}}+C.$