# a special case of partial integration

In determining the antiderivative of a transcendental (http://planetmath.org/AlgebraicFunction) function $U$ whose derivative $U^{\prime}$ is algebraic (http://planetmath.org/AlgebraicFunction), the result can be obtained when choosing in the formula

 $\int UV^{\prime}\,dx=UV-\!\int VU^{\prime}\,dx$

of integration by parts$V^{\prime}\equiv 1$;  then one has

 $\int U\,dx=\int U\!\cdot\!1\,dx\;=\;U\!\cdot\!x-\!\int x\!\cdot\!U^{\prime}\,dx.$

The functions $U$ in question are mainly the logarithm (http://planetmath.org/NaturalLogarithm2), the cyclometric functions and the area functions.

Examples.

1. 1.

$\displaystyle\int\!\ln{x}\,dx=x\ln{x}-\!\int x\!\cdot\!\frac{1}{x}\,dx=\;x\ln{% x}-x+C$

2. 2.

$\displaystyle\int\!\arcsin{x}\,dx=x\arcsin{x}-\!\int\!x\!\cdot\!\frac{1}{\sqrt% {1\!-\!x^{2}}}\,dx=x\arcsin{x}+\frac{1}{2}\int\!\frac{-2x}{\sqrt{1\!-\!x^{2}}}% \,dx\\ =x\arcsin{x}+\sqrt{1\!-\!x^{2}}+C$

3. 3.

$\displaystyle\int\!\arctan{x}\,dx=x\arctan{x}-\!\int\!x\!\cdot\!\frac{1}{1\!+% \!x^{2}}\,dx=x\arctan{x}-\!\frac{1}{2}\int\!\frac{2x}{1\!+\!x^{2}}\,dx\\ =x\arctan{x}-\frac{1}{2}\ln(1\!+\!x^{2})+C=\;x\arctan{x}-\ln\sqrt{1\!+\!x^{2}}+C$

4. 4.

$\displaystyle\int\!\operatorname{arcosh}{x}\,dx=x\operatorname{arcosh}{x}-\!% \int\!x\!\cdot\!\frac{1}{\sqrt{x^{2}\!-\!1}}\,dx\\ =x\operatorname{arcosh}{x}-\sqrt{x^{2}\!-\!1}+C$

The choice  $V^{\prime}\equiv 1$  works as well in such cases as  $\int(\ln{x})^{2}\,dx$  and  $\int\ln(\ln{x})\,dx$,  giving respectively  $x((\ln{x})^{2}-2\ln{x}+2)+C$  and  $x\ln(\ln{x})-\operatorname{Li}{x}+C$ (see logarithmic integral). Also  $\int(\arcsin{x})^{2}\,dx$  , requiring two integrations by parts, and giving the result  $x(\arcsin{x})^{2}+2\sqrt{1\!-\!x^{2}}\arcsin{x}-2x+C$.

Title a special case of partial integration ASpecialCaseOfPartialIntegration 2013-03-22 17:38:35 2013-03-22 17:38:35 pahio (2872) pahio (2872) 11 pahio (2872) Feature msc 26A36 IntegralTables