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How do you integrate #1/(sqrt(25+4x^2))# ?

Answer 1

I would take out, from the square root, #25# getting:

#int1/(5*sqrt(1+4/25x^2))dx=# #=int1/5*1/(sqrt(1+(2/5x)^2))dx=# #=1/5*5/2sinh^(-1)(2/5x)+c# #=1/2sinh^(-1)(2/5x)+c#

Where #sinh^(-1)(x)=#arcsinh(x) is one of the inverse hyperbolic functions . Have a look at: https://tutor.hix.ai for more info.

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Answer 2

Emilia Aguilar

To integrate ( \frac{1}{\sqrt{25 + 4x^2}} ), you can use the substitution method. Let ( u = 5x ), then ( du = 5 dx ). This implies ( dx = \frac{du}{5} ). Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{1}{\sqrt{25 + 4x^2}} dx = \frac{1}{5} \int \frac{1}{\sqrt{1 + u^2}} du ]

This is now a standard integral of ( \frac{1}{\sqrt{1 + u^2}} ), which is the inverse hyperbolic sine function, (\text{arcsinh}(u)), also denoted as ( \sinh^{-1}(u) ). So,

[ \int \frac{1}{\sqrt{25 + 4x^2}} dx = \frac{1}{5} \sinh^{-1}(u) + C ]

Finally, substitute back ( u = 5x ) to get the final result:

[ \int \frac{1}{\sqrt{25 + 4x^2}} dx = \frac{1}{5} \sinh^{-1}(5x) + C ]

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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.