Question

∫(4supra √ x^2-16+8 supra x^2+16) dx ;x∈(4,+∞),

∫(6 supra x^2-9+4 supra √x^2+25)dx;x∈(-∞,-3)

Answer 1

$\star) \int \frac{4}{ \sqrt{ {x}^{2} - 16} } + \frac{8}{ \sqrt{ {x}^{2} + 16} } \: dx$

$= \int \frac{4}{ \sqrt{ {x}^{2} - 16} } \: dx + \int \frac{8}{ \sqrt{ {x}^{2} + 16} } \: dx$

$= 4 \int \frac{1}{ \sqrt{ {x}^{2} - 16 } } \: dx + 8 \int \frac{1}{ \sqrt{ {x}^{2} + 16} } \: dx$

$= 4ln(x + \sqrt{ {x}^{2} - 16} ) + 8ln(x + \sqrt{ {x}^{2} + 16} ) + C$

$\star) \int \frac{6}{ {x}^{2} - 9} + \frac{4}{ \sqrt{ {x}^{2} + 25 } } \: dx$

$= \int \frac{6}{ {x}^{2} - 9} \: dx + \int \frac{4}{ \sqrt{ {x}^{2} + 25 } } \: dx$

$= 6 \int \frac{1}{ {x}^{2} - {3}^{2} } \: dx + 4 \int \frac{1}{ \sqrt{ {x}^{2} + 25} } \: dx$

$= 6 \times \frac{1}{2 \times 3} ln | \frac{x - 3}{x + 3} | + 4ln(x + \sqrt{ {x}^{2} + 25} ) + C$

$= ln | \frac{x - 3}{x + 3} | + 4ln(x + \sqrt{ {x}^{2} + 25 } ) + C$