by Guest » Sat Jul 04, 2020 8:27 am
From the formula: [tex]\int \frac{dx}{\sqrt{a^2-x^2}}=Arcsin \frac{x}{a}+C[/tex]
So we have, [tex]\int \frac{dx}{\sqrt{2-x^2}}=Arcsin \frac{x}{\sqrt{2}}+C[/tex]
For [tex]\int \frac{dx}{2x^{2}-6}[/tex], you can factor out 2, so you have,
[tex]\frac{1}{2} \int \frac{dx}{x^{2}-3} = \frac{1}{2} \int \frac{dx}{(x - \sqrt{3})(x+\sqrt{3})}[/tex]
Then use partial fractions:
[tex]= \frac{1}{2} \int \frac{dx}{(2 \sqrt{3})(x+\sqrt{3})} - \frac{1}{2} \int \frac{dx}{(2 \sqrt{3})(x - \sqrt{3})}[/tex]
[tex]=\frac{1}{4 \sqrt{3}} ( \ln \mid {x+ \sqrt{3}} \mid - \ln \mid {x - \sqrt{3}} \mid ) +C[/tex]
[tex]=\frac{1}{4 \sqrt{3}} \ln \mid \frac { {x+ \sqrt{3} } } { {x - \sqrt{3} } } \mid +C[/tex]
depending if its ok that there is a radical in the denominator..