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

$$\int U{V}^{\prime }𝑑x=UV-\int V{U}^{\prime }𝑑x$$

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

$$\int U𝑑x=\int U\cdot 1𝑑x=U\cdot x-\int x\cdot U^{\prime }𝑑x.$$

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

Examples.

1. ${\displaystyle \int \ln xdx}=x\ln x-{\displaystyle \int x\cdot \frac{1}{x}𝑑x}=x\ln x-x+C$

2. ${\displaystyle \int \arcsin xdx}=x\arcsin x-{\displaystyle \int x\cdot \frac{1}{\sqrt{1-x^2}}𝑑x}=x\arcsin x+{\displaystyle \frac{1}{2}}{\displaystyle \int \frac{-2x}{\sqrt{1-x^2}}𝑑x}=x\arcsin x+\sqrt{1-x^2}+C$

3. ${\displaystyle \int \arctan xdx}=x\arctan x-{\displaystyle \int x\cdot \frac{1}{1+x^2}𝑑x}=x\arctan x-{\displaystyle \frac{1}{2}}{\displaystyle \int \frac{2x}{1+x^2}𝑑x}=x\arctan x-{\displaystyle \frac{1}{2}}\ln (1+x^2)+C=x\arctan x-\ln \sqrt{1+x^2}+C$

4. ${\displaystyle \int \mathrm{arcosh}xdx}=x\mathrm{arcosh}x-{\displaystyle \int x\cdot \frac{1}{\sqrt{x^2-1}}𝑑x}=x\mathrm{arcosh}x-\sqrt{x^2-1}+C$

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