∫√3+2xdx

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Is 3+2x all under the square root or is it just root 3?
\[\sqrt{3+2x}\] ?
yes

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Other answers:

Ok. We will use u-substitution. Set u equal to what's under the square root (3+2x) and then take the derivative of 3+2x to find du. So du = 2 dx
ok
So you are left with \[\int\limits \sqrt{u}/2 du\]
The reason we divide by 2 is when you get du = 2 dx you want dx by itself, so you get du/2 = dx
ok
Since your dividing by 2, you can pull a 1/2 out in front of the integral, leaving you with the integral of root u, which can be rewritten as u^(1/2). Then to integrate we add 1 to the exponent (giving us u^(3/2)). We have to multiply by the reciprocal of our new exponent.
So your final answer should be [2u^(3/2)]/3
ok
Since 2/3 times u^(3/2) just moves the u^(2/3) on top and multiplies it by 2.
ok
Any other questions?
yes
Ok what?
∫x+1/2x-x2+2 dx
Oh well I meant about this problem. I'll look at it and see if I can help in the other problem.
tthanks God bless you

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