Integrate dx/2-x^2 =?

1 Answer
Apr 26, 2018

=1/sqrt2ln(sqrt2/sqrt(2-x^2)+x/sqrt(2-x^2))+C

Explanation:

intdx/(2-x^2)

Using Trigonometric substitution

x=sqrt2sinu

dx=sqrt2cosu*du

Substitute

int(sqrt2cosu*du)/(2-2sin^2u)

color(green)(2-2sin^2u=2cos^2u)

=int(sqrt2cosudu)/(2cos^2u

Simplify

=1/sqrt2int(du)/cosu

color(green)(1/cosu=secu)

=1/sqrt2intsecudu

=1/sqrt2ln(secu+tanu)+C

Reverse the Substitution

=1/sqrt2ln(sqrt2/sqrt(2-x^2)+x/sqrt(2-x^2))+C