How to integrate int (x^2 + 4)^(-1/2) dx Is it a substitution?

AI Thread Summary
To integrate (x^2 + 4)^(-1/2) dx, a trigonometric substitution is recommended, utilizing a right triangle where the hypotenuse is the radical and one side is 2. By letting u = x^2 + 4, the differential du can be expressed as du = 2x dx, leading to dx = du/(2x). Substituting these values simplifies the integral to (1/2) ∫ u^(-1/2) du, which results in (x^2 + 4)^(3/2) + C after integration. This method effectively resolves the integration challenge presented.
gordda
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I need help antidiffing this equation:
(x^2+4)^(-1/2)

i have tried subbing u=x^2+4 i have tried subbing u= (x^2+4)^(-1/2).
i have tried making x the subject. even tried to use partial fraction, with no avail, because i could not figure out how to use partially factorize it.
If anyone could lead me towards the right answer.
Thanks.
 
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gordda said:
I need help antidiffing this equation:
(x^2+4)^(-1/2)

i have tried subbing u=x^2+4 i have tried subbing u= (x^2+4)^(-1/2).
i have tried making x the subject. even tried to use partial fraction, with no avail, because i could not figure out how to use partially factorize it.
If anyone could lead me towards the right answer.
Thanks.


Whenever you have an integrand that looks like the Pathagoras theorem, use trig substitutions. You know, draw a right triangle with theta in it. Then the hypotneuse is the radical, one side is 2, the other is x right. Follow through: What would the tan(theta) be?
 


Yes, integrating (x^2+4)^(-1/2) can be done using substitution. Here's how:

Let u = x^2 + 4
Then du/dx = 2x
And dx = du/2x

Substituting these values into the original equation, we get:

∫ (x^2+4)^(-1/2) dx = ∫ (x^2+4)^(-1/2) * (du/2x)
= (1/2) * ∫ u^(-1/2) du
= (1/2) * ∫ u^(-1/2) * u^0 du
= (1/2) * ∫ u^(1/2) du
= (1/2) * (2u^(3/2)) + C
= u^(3/2) + C
= (x^2+4)^(3/2) + C

So, the final integrated form is (x^2+4)^(3/2) + C. You can also verify this solution by differentiating it and checking if it gives the original equation.

Hope this helps!
 
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