Solve ∫dx/(x(√(25-x²)))?

Question

Im not sure what I'm supposed to substitute in!

Answer 1

Try this site and work backwards.

I'm thinking u = 25 - x^2
or u^2 = 25 - x^2

Answer 2

Q1: should be 2 & -a million once you ingredient Q2: your answer for C is the critical of dC/dx so it really is going to have a consistent time period on the best, or +D (i picked D particularly of c for far less confusion hence C(0) = D = 2300 in view that once you plug 0 in the in hassle-free words project that continues to be is the D time period. so then the answer is: C = (4x^5/4)/5 ln 10 + 10x + 2300

Answer 3

Let u = √(25-x^2)
u^2 = 25-x^2
2udu = -2xdx

∫dx/(x(√(25-x²)))
= ∫xdx/[x^2(√(25-x²)]
= ∫-udu/[(25-u^2)u]
= ∫du/(u^2-5^2)
= (1/10)∫[1/(u-5) - 1/(u+5)]du
= (1/10)[ln|u-5| - ln|u+5|]+c
= (1/10)[ln|√(25-x²)-5| - ln|√(25-x²)+5| ]+c

Answer 4

Solve that? Sorry, no thank you. I have an essay I am strategically avoiding that I would rather write. And I don't think you'd appreciate the incorrect answer I would come up with if I did attempt to solve it. Good luck, I hope someone else has some idea?

Answer 5

Let x = 5 sin u
dx = 5 cos u du
I = 5 ∫ cos u du / (5sinu).√[(25 - 25 sin² u]
I = ∫ cos u du / [ sin u √(25 - 25 sin² u) ]
I = ∫ cos u du / [ sin u √ [ 25(1 - sin² u) ]
I = (1/5).∫ cos u du / (sin u cosu)
I = (1/5).∫ (1 / sin u).du
I = (1/5).∫ cosec u du
I = (1/5).log (tan (u /2)) + C
I = (1/5) log tan (sin^(-1) (x /10) ) + C

Answer 6

I go with Como myself. It was what I worked out without carrying out full integration as I had to take my kid to school

Answer 7

I know the answer but i'm not telling you,..............not lol
You have to learn for yourself.

;-)

Answer 8

Your question is not clear.

What is the little squiggle? What are you supposed to do?

Answer 9

dxfx

Answer 10

No Thanks - I'm not sure either.