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How do you find the Integral of #dx/sqrt(x^2+16)#?

Answer 1

Elijah Abram

#int (dx)/ sqrt(x^2+16) dx = int 1/(4sqrt((x/4)^2+1)) dx#

#=int 1/(sqrt((x/4)^2+1))( 1/4)dx #

#u = x/4#, #du=1/4 dx# and #int 1/sqrt(u^2+1) du=sinh^(-1) u +C#

So #int (dx)/ sqrt(x^2+16) dx = sinh^(-1) (x/4) +C#

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Answer 2

Adam Adkins

To find the integral of dx/sqrt(x^2+16), you can use the trigonometric substitution method. Let x = 4tan(θ), then dx = 4sec^2(θ)dθ. Substitute these expressions into the integral, and simplify it using trigonometric identities. You will end up with the integral of sec(θ)dθ, which can be evaluated easily. Finally, replace θ with its corresponding expression in terms of x to obtain the final result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.