Question

how do I integrate 1/(0.0625-0.00016x^2) ???

Answer 1

For a function like this, write
\[f(x) = { 1 \over a^2 - b^2x^2 } = \frac{1}{2a} \left({1 \over a - bx} + {1 \over a + bx}\right)\]
then the integral is a sum of terms in ln.

Answer 2

See what to do now?

Answer 3

(If I had to guess, I'm guessing the second term in 0.0016x^2; one less zero than you wrote.)

Answer 4

still there?

Answer 5

I agree with James. it's much nicer with .0016x^2

Answer 6

it says 16^(-5)

Answer 7

You're calculating a definite integral? Even if you are, you're going to have to calculate the indefinite integral first.

Answer 8

\[\int\limits_{}^{}\frac{1}{(\frac{25}{100})^2-\frac{16}{100000} x^2} dx\]
\[\int\limits_{}^{}\frac{1}{(\frac{25}{100})^2-(\frac{4}{100})^2 \cdot \frac{1}{10} x^2}\]
\[\frac{1}{\frac{1}{100^2}}\int\limits_{}^{}\frac{1}{(25)^2-(\frac{4x}{\sqrt{10}})^2} dx\]
\[100^2 \int\limits_{}^{}\frac{\sqrt{10}^2}{(25)^2 (\sqrt{10}^2)-4x^2} dx\]
\[100^2 \cdot 10 \int\limits_{}^{}\frac{1}{25^2\sqrt{10^2}-4x^2} dx\]
i would write like this
and would use a trig substitution (but this doesn't mean you have to)

Answer 9

i hate decimals
they are yucky

Answer 10

oops 4 squared

Answer 11

\[100^2 \cdot 10 \int\limits_{}^{}\frac{1}{(25 \cdot \sqrt{10})^2-4^2x^2} dx\]

Answer 12

Alternatively, in my notation a = 0.25 = 1/4 and b = 0.04 = 1/25

Now\[\int\limits \frac{dx}{a^2 - b^2x^2} = \frac{1}{2a} \int\limits \left( \frac {1}{a+bx} + \frac{1}{a-bx} \right) dx = \frac{1}{2ab} (\ln(a+bx) - \ln(a-bx))\]