What is the antiderivative of #sqrt(4x² + 1)#?

Answer 1

#= 1/4 ( 2x sqrt{1+ 4x^2})+ sinh^{-1} 2x) + C#

using the identity #cosh^2 t - sinh^2 t =1 # i'd got with a sub #4x^2 = sinh^2 t # ie #2x = sinh t #, so # 2 \ dx = cosh t \ dt #

that leaves:

#int \ sqrt(4x² + 1) \ dx = int \ sqrt{sinh^2 t + 1} \ 1/2 cosh t \ dt#

# = 1/2 int \ cosh^2 t \ dt#

we use the following hyperbolic identities

#cosh^2 t = 1/2 (cosh 2t + 1)#

so the integral becomes

# = 1/4 int \ cosh 2t + 1 \ dt#

#= 1/4 (1/2 sinh 2t + t) + C#

within this #t = sinh^{-1} 2x# from above

for #sinh 2t#, look at the above identities again.

#sinh 2t = 2 sinh t cosh t #
#= 2 sinh t sqrt{1 + sinh^2 t)#
# = 4x sqrt{1+ 4x^2}#

so

# 1/4 (1/2 sinh 2t + t) + C #

#= 1/4 (1/2 * 4x sqrt{1+ 4x^2} + sinh^{-1} 2x) + C#

#= 1/4 ( 2x sqrt{1+ 4x^2})+ sinh^{-1} 2x) + C#

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Answer 2

The antiderivative of ( \sqrt{4x^2 + 1} ) with respect to ( x ) is ( \frac{1}{2} \left( x \sqrt{4x^2 + 1} + \frac{1}{2} \ln |2x + \sqrt{4x^2 + 1}| \right) + C ), where ( C ) is the constant of integration.

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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.

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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