integrate 1/[root(2x^2+x-3) ] dx
Urgent
∫ 1/√(2x^2+x-3) dx = 1/√2 ∫ 1/√((x + 1/4)^2 - (5/4)^2) dx
Now, if you let y = x + 1/4, that is just
1/√2 ∫ 1/√(u^2 - (5/4)^2) du
and don't you have a standard form for
∫ du/√(u^2 - a^2) ?
Urgent
Now, if you let y = x + 1/4, that is just
1/√2 ∫ 1/√(u^2 - (5/4)^2) du
and don't you have a standard form for
∫ du/√(u^2 - a^2) ?