HELP! how can i integrate: 1 / ( x^2 ( 4 + x^2)^1/2 )?

Answer 1

Let x = 2tanθ
So dx = 2sec²θ dθ

∫1 / (x² √[4+x²]) dx
=∫1 / (4tan²θ √[4+4tan²θ]) 2sec²θdθ
=∫1 / (4tan²θ √[4sec²θ]) 2sec²θ dθ
=∫1 / (8tan²θ secθ) 2sec²θ dθ
=1/4∫secθ / (tan²θ) dθ
=1/4∫cosθ / sin²θ dθ
Let u = sinθ
du = cosθ dθ
So 1/4∫cosθ / sin²θ dθ
= 1/4∫1 / u² du
= -1/4u + c
= -1/4sinθ + c
Now tanθ = x / 2 so sinθ = √[4+x²] / x
So the integral is equal to
- √[4+x²] / 4x + c

Answer 2

Calculus textbooks usually provide a list of integrals. Look at the sub-list for "forms containing (u^2 +/- a^2).

The one that is appropriate for our specific problem is:
∫du/(u^2(u^2 + a^2)^1/2) = -((u^2 + a^2)^1/2)/(a^2)u

So, here we have
x^2 = u^2
x = u
4 = a^2

Plug in these values, the answer is:
-((x^2 + 4)^1/2)/4x

Check by getting the derivative of our answer.

Answer 3

use the substitution --> x= 2 tanA
put in the above equation, you get 1/(4 tan A^2 (2 sec A))
--> 1/(4 tan A^2 * 2 sec A)


No use, sec A = t
do simple substitution now to obtain the answer :)

Best of Luck!

Answer 4

Use a trigonometric substitution.

Answer 5

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