Question

integral de cos^-1 2x dx por partes

Answer 1

Respuesta:

$x\arccos \left(2x\right)-\frac{1}{2}\sqrt{1-4x^2}+C$

Explicación:

$\int\limits^ {} cos^{-1} 2x\, dx \\\int\limits^ {} arccos 2x\, dx \\u = arccos 2x\\u' = \frac{-2}{\sqrt{1-4x^{2} } } \\v' = 2x\\v = x^{2} +C\\\\x\arccos \left(2x\right)-\int \:-\frac{2x}{\sqrt{1-4x^2}}dx\\\\x\arccos \left(2x\right)-\frac{1}{2}\sqrt{1-4x^2}+C$