solve the integral of 1/(x^2-1/4)?

Question

explain work please

Answer 1

wait is is 1/((x^2)-14) or 1/((x^(2-14))

Answer 2

The denominator can be factored:

x^2 - 1/4 = (x - 1/2) (x + 1/2)

That means that the fraction 1/(x^2 - 1/4) can be split in two terms,

1/(x^2 - 1/4) = A/(x - 1/2) + B/(x + 1/2)

You find A and B by solving
1 = A(x + 1/2) + B(x - 1/2)
It turns out that A = +1 and B = -1, so we have

1/(x^2 - 1/4) = 1/(x - 1/2) - 1/(x + 1/2)

These two terms can be integrated separately. Their integrals are:

ln |x - 1/2| and - ln |x + 1/2| (plus a constant, of course)

So the final answer is
ln |x - 1/2| - ln |x + 1/2| + C

Answer 3

You split the fraction via partial fractions first.

1/(x^2-1/4) = 1/[(x+1/2)(x-1/2)]
= A/(x+1/2)+B(x-1/2)

By equating numerators, 1 = A( x - 1/2) + B(x + 1/2)
Comparing x^2 term on both sides, 0 = A+B
Comparing constants on both sides, 1 = B/2 - A/2
Solving for A and B, we get A= -1, B= 1

Integrate and we get:
-ln|x+1/2| + ln|x-1/2| + c

Answer 4

Integrate 1/(x^2 - 1/4) = intergrate 1/(x^2 - (1/2)^2)
=1/(1/2) tan^-1 ( x / (1/2))
= 2 tan^-1 (2x)

Answer 5

U've 2 ways: first:
1/((x-0.5)(x-+0.5))= A/(x-0.5)+B/(x+0.5) and A=1 and B=-1, then ur integral becomes: 1/(x-0.5)+1/(x+0.5) = ln(C*(x-0.5)/(x+0.5)).

Second:
Int(1/(x^2-a^2))=(1/a) * Intanh(x/2) + C; hence ur integral is equal to 2*Intanh(2x)+C.

Answer 6

1/(x^2- 1/4)
= 1/ x^2 - (1/2)^2
=1/(x+ 1/2) (x- 1/2)
(use a^2 - b^2 = (a+b) (a-b) )

so this is the ans.
but the first answerer was right. ushudve put a () on 1/4 or it gets confusing.
But i gess disis wat u ment 2say.

Answer 7

1/x^2-1/4=4/4x^2-1
2x=cosz,[tanz=1-4x^2]
2dx=-sinz*dz
dx=-1/2*sinz*dz
I=-int(4/cos^2z-1)*1/2*sinz*dz
=-int[2sinz*dz/sinz^2]
=-int[2dz/sinz]
=-int[2cosecz.dz]=-2*ln tanz/2
=-2ln tan[-2/i-4x^2]
=2lntan[2/1-4x^2]

Answer 8

Explain the parenthesis, or write it clearly please