Question

Integrate dx/(25-16x^2)

Answer 1

No one?

Answer 2

Never mind, I made trough the question. :)

Answer 3

\[\large\int\frac{dx}{a^2 - (x)^2} = \frac 1 {2a}ln(\frac{x+a}{x-a}) \]

Can/did you use this?

Sorry that I took so long to type a reply.

Answer 4

'+ c' --> constant of integration, ofcourse.

Answer 5

I'll turn on my pc to answer you, wait a sec

Answer 6

try this..\[4x=5sinu\]

Answer 7

Nope, can't use that.
I used Integration by substitution:

First i had to ajust my equation:
\[\int\limits_{?}^{?} dx/\sqrt{25-16x ^{2}} = 1/5\int\limits_{?}^{?} dx/\sqrt{1-16x^{2}/25}\]

Ready to go:
\[1/5\int\limits_{}^{}dx/\sqrt{1-16x ^{2}/25}\]
\[u=4x/5\]
\[du=4/5\]
\[1/4\int\limits\limits_{}^{} dx/\sqrt{1-u ^{2}}\]
\[\int\limits\limits_{}^{} dx/\sqrt{25-16x ^{2}} = (1/4)\sin ^{-}(4x/5) + C\]

Answer 8

I solved it in that way..