# Integrate the following #int sqrt(x^2+81) dx#?

This is a relatively known integral. The proof can be found here.

Hopefully this helps!

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To integrate ∫√(x^2 + 81) dx, we use trigonometric substitution. Let x = 9tan(θ), then dx = 9sec^2(θ) dθ. Substituting these into the integral gives:

∫√(x^2 + 81) dx = ∫√(81tan^2(θ) + 81) * 9sec^2(θ) dθ = ∫√(81(sec^2(θ))) * 9sec^2(θ) dθ = ∫√(81sec^2(θ)) * 9sec^2(θ) dθ = ∫9sec(θ) * 9sec^2(θ) dθ = ∫81sec^3(θ) dθ.

Now, we integrate ∫sec^3(θ) dθ using the reduction formula for integrals of secant functions:

∫sec^n(θ) dθ = (1/(n-1))sec^(n-2)(θ)tan(θ) + (n-2)/(n-1) ∫sec^(n-2)(θ) dθ.

Using this formula with n = 3, we have:

∫sec^3(θ) dθ = (1/2)sec(θ)tan(θ) + (1/2)∫sec(θ) dθ.

The integral of sec(θ) is a standard integral, which is ln|sec(θ) + tan(θ)|. So, substituting this back into our expression gives:

∫sec^3(θ) dθ = (1/2)sec(θ)tan(θ) + (1/2)ln|sec(θ) + tan(θ)| + C.

Now, we substitute back the expression for θ in terms of x, which was θ = arctan(x/9):

= (1/2)(x/√(x^2 + 81))(√(x^2 + 81)) + (1/2)ln|x + √(x^2 + 81)| + C = (1/2)x + (1/2)ln|x + √(x^2 + 81)| + C.

So, the integral of √(x^2 + 81) dx is (1/2)x + (1/2)ln|x + √(x^2 + 81)| + C, where C is the constant of integration.

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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.

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.

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.

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