Verified answer

∫2x√(6x^2-1)dx

Let u = 6x^2-1

du = 12x dx

x dx = du/12

2x dx = du/6

∫2x√(6x^2-1)dx = (1/6) ∫ √u du

= (1/6) ∫ u^(1/2) du

= (1/6) u^(1/2+1) /(1/2+1)

= (1/6)(2/3) u^(3/2) + C

= (1/9) (6x^2-1)^(3/2) + C

= (6x^2-18)√(6x^2-18)/9 + C

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Note: x^(3/2) = x √x

enable's do ? arcsin x dx first. and enable's try this a touch distinctive way. enable u = arcsin x, x = sin u, dx = cos u du. So the vital will become ? u cos u du. we are able to try this via factors, yet I fairly want tabular integration. u cos u a million sin u 0 -cos u.s. ? u cos u du = u sin u + cos u + C. Now we translate each and everything lower back to x and get ? arcsin x dx = x arcsin x + ?(a million-x^2) + C. So the respond on your venture is 9(x arcsin x + ?(a million-x^2)) + C