In determining the antiderivative of a transcendental function $U$ whose derivative $U'$ is algebraic, the result can be obtained when choosing in the formula $$\int UV'\,dx = UV-\!\int VU'\,dx$$ of integration by parts $V' \equiv 1$ ; then one has $$\int U\,dx = \int U\!\cdot\!1\,dx\; =\; U\!\cdot\!x-\!\int x\!\cdot\!U'\,dx.$$ The functions $U$ in question are mainly the logarithm, the cyclometric functions and the area functions.

Examples.

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

The choice $V' \equiv 1$ 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 $x\ln(\ln{x})-\Li{x}+C$ (see logarithmic integral). Also $\int(\arcsin{x})^2\,dx$ succeeds, requiring two integrations by parts, and giving the result $x(\arcsin{x})^2+2\sqrt{1\!-\!x^2}\arcsin{x}-2x+C$ .