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What is the antiderivative with respect to x of y=(x^2+2x+2)^(1/2)?
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The antiderivative of a function is equivalent of the integral of that function.
Therefore the antiderivative of `(x^2+2x+2)^(1/2)` with respect to `x` is equivalent to
Rewriting `x^2 + 2x + 2` as `(x+1)^2 + 1` this is equal to
`int [(x+1)^2 + 1]^(1/2)dx`
Now make the substitution
`tan(t) = x+1` , where `(dx)/(dt) = sec^2t` . The integral is then equal to
`int (tan^2t + 1)^(1/2)sec^2t dt`
Since `tan^2t +1 = sec^2t` (trigonometric identity) this equals
`int sec^3t dt `
`= 1/2 sec(t)tan(t) + 1/2ln|sec(t) +tan(t)| + c`
`= 1/2sqrt(x^2+2x+2)(x+1) + 1/2ln|sqrt(x^2+2x+2) + (x+1)| + c`
(since `sec(arctan(x+1)) = sqrt((x+1)^2+1) = sqrt(x^2+2x+2)` )
This result is found by using integration by parts where we let
`u = sec(x)` and `v = tan(x)` and solve
`int u (dv)/(dx) = uv - int v (du)/(dx)`
The antiderivative of (x^2+2x+2)^(1/2) is equivalent to the integral of the same expression. Using substitution and then integration by parts this is found to be
1/2(x^2+2x+2)^(1/2)(x+1) + 1/2 ln|(x^2+2x+2)^(1/2) + (x+1)| + c
Posted by mathsworkmusic on May 27, 2013 at 3:29 PM (Answer #3)
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