find the following indefinite integral?

Hi

Use the substitution u = 1 - 3x^2, du = -6x dx.

∫ -2x/√(1 - 3x^2) dx

= 1/3 ∫ -6x/√(1 - 3x^2) dx

= 1/3 ∫ 1/√(1 - 3x^2) * -6x dx

= 1/3 ∫ 1/√u du

= 1/3 ∫ u^(-1/2) du

= (1/3)*2u^(1/2) + C

= (2/3)(√u) + C

= (2/3)√(1 - 3x^2) + C

I hope this helps!

2x / sqrt(a million - 3x^2) dx i visit ingredient the two exterior of the crucial. subsequently our integrand (the element to the appropriate of the crucial sign), is only: x / sqrt(a million - 3x^2) dx Say that: u = a million - 3x^2 Calculate the derivative of u: du = -6x * dx Now if we'd pick to replace u into our integrand, we ought to verify we also have a corresponding fee of du. We at the instant purely have x*dx. So if we multiply the integrand through -6, we will have -6x*dx, it relatively is comparable to du. yet, we can not only manage one a million/2, we ought to the two and oppositely do a similar on the different section. subsequently we multiply through -a million/6 on the exterior. If we've been to combine this, it may be comparable to multiplying the equation through -6/-6, or only +a million (which, as all of us understand, does not replace the value in any respect): -2/6 * ? du/sqrt(u) At this factor, you may, in case you pick, replace back in du = -6xdx and u=a million-3x^2 to verify that i did not basically replace the subject in any respect. Now we are in a position to calculate a million/sqrt(u)'s antiderivative certainly with a potential rule, understanding that it relatively is only u^(-a million/2). subsequently we get: 2u^(a million/2) + c the place c is an arbitrary consistent of integration. Then you ought to multiply out the scalar -2/6 we had in front and resubstitute u = a million - 3x^2: (-a million/3)(2(a million - 3x^2)^(a million/2)) + c this is not significant we multiply c through the -a million/3 as we are in a position to %. any huge style for c and get a valid answer. Simplifying: (-2/3) * sqrt(a million - 3x^2) + c

let u = 1 - 3x²

du = -6x dx

You should be able to do it from there. If not, go learn about u-substitutions. I'm not here to do your homework.