Question

ayuda es muy importante mi calificación depende de esto :c

Answer 1

Respuesta:

1)  $\int \frac{x^2}{\sqrt{x^2-6}}dx$

=$6\left(\frac{1}{12}x\sqrt{x^2-6}+\frac{1}{2}\ln \left|\sqrt{\frac{1}{6}\left(x^2-6\right)}+\frac{1}{\sqrt{6}}x\right|\right)+C$

2)$\int \frac{x^2}{\sqrt{x^2+5}}dx=5\left(-\frac{1}{2}\ln \left|\frac{1}{\sqrt{5}}x+\sqrt{\frac{1}{5}\left(5+x^2\right)}\right|+\frac{1}{10}x\sqrt{5+x^2}\right)+C$

Explicación:

1

$=\int \:6\sec ^3\left(u\right)du$

$=6\left(\frac{\sec ^2\left(u\right)\sin \left(u\right)}{2}+\frac{1}{2}\cdot \int \sec \left(u\right)du\right)$

$=6\left(\frac{\sec ^2\left(u\right)\sin \left(u\right)}{2}+\frac{1}{2}\ln \left|\tan \left(u\right)+\sec \left(u\right)\right|\right)$

$=6\left(\frac{\sec ^2\left(\arcsec \left(\frac{1}{\sqrt{6}}x\right)\right)\sin \left(\arcsec \left(\frac{1}{\sqrt{6}}x\right)\right)}{2}+\frac{1}{2}\ln \left|\tan \left(\arcsec \left(\frac{1}{\sqrt{6}}x\right)\right)+\sec \left(\arcsec \left(\frac{1}{\sqrt{6}}x\right)\right)\right|\right)$$=6\left(\frac{1}{12}x\sqrt{x^2-6}+\frac{1}{2}\ln \left|\sqrt{\frac{1}{6}\left(x^2-6\right)}+\frac{1}{\sqrt{6}}x\right|\right)$

te resumí algunos pasos