Question

Vreau să ştiu dacă am rezolvat corect :

$\star) \int \frac{18}{3 {x}^{2} + 27} \: dx,x \: \in \: \mathbb{R}$

$= \int \frac{18}{3( {x}^{2} + 9)} \: dx = 6 \int \frac{1}{ {x}^{2} + {3}^{2} } \: dx$

$= 6 \times \frac{1}{3} \times arctg \: \frac{x}{3}$

$= 2 \: arctg \: \frac{x}{3} + C$

$\star) \int( {5}^{x} ln5 - {4}^{x} ln16)dx \: ,x\in\mathbb{R}$

$= ln5 \int {5}^{x} \: dx - ln16 \int {4}^{x} \: dx$

$= ln5 \times \frac{ {5}^{x} }{ln5} - ln16 \times \frac{ {4}^{x} }{ln4}$

$= {5}^{x} - {4}^{x} ln12 + C$

$\star) \int \frac{1}{ \sqrt{6 {x}^{2} + 24} } \: dx,x\in\mathbb{R}$

$= \int \frac{1}{ \sqrt{6( {x}^{2} + 4)} } \: dx = \frac{1}{ \sqrt{6 } } \int \frac{1}{ \sqrt{ {x}^{2} + {2}^{2} } } \: dx$

$= \frac{ \sqrt{6} }{6} \: ln(x + \sqrt{ {x}^{2} + {2}^{2} )} + C$

$\star) \int \frac{1}{ \sqrt{ {2x}^{2} - 18 } } \: dx \: ,x>3$

$= \int \frac{1}{ \sqrt{2( {x}^{2} - 9) } } \: dx = \frac{1}{ \sqrt{2} } \int \frac{1}{ \sqrt{ {x}^{2} - {3}^{2} } } \: dx$

$= \frac{ \sqrt{2} }{2} \: ln(x + \sqrt{ {x}^{2} - {3}^{2} } ) + C$

$\star) \int \frac{ \sqrt{3} }{ \sqrt{48 - {3x}^{2} } } \: dx \: ,x\in(-2,2)$

$= \int \frac{ \sqrt{3} }{ \sqrt{3(16 - {x}^{2} )} } \: dx = \frac{ \sqrt{3} }{ \sqrt{3} } \int \frac{1}{ \sqrt{16 - {x}^{2} } } \: dx$

$= \int \frac{1}{ \sqrt{ {4}^{2} - {x}^{2} } } \: dx = arcsin \: \frac{x}{4} + C$

Answer 1

da este foarte corecta ca am facuto si eu la tabla si chiar asa am scris te rog sa ma crezi