Evaluate ∫ (1-x^2)/(x- 2x^2) dx?

Since the numerator and denominator have the same degree, perform division to get:

(1 - x^2)/(x - 2x^2) = 1/2 + [1 - (1/2)x]/(x - 2x^2).

Then, apply partial fractions to [1 - (1/2)x]/(x - 2x^2) to get:

[1 - (1/2)x]/(x - 2x^2) = A/x + B/(1 - 2x), for some A and B.

(Note that the denominator factors to x(1 - 2x).)

Multiplying both sides by x - 2x^2 gives:

1 - (1/2)x = A(1 - 2x) + Bx.

(i) Letting x = 0:

1 - 0 = A(1 - 0) ==> A = 1

(ii) Letting x = 1/2:

1 - (1/2)(1/2) = B/2 ==> B = 3/2.

Thus:

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

= 1/2 + 1/x + (3/2)/(1 - 2x).

Integrating term-by-term yields:

∫ (1 - x^2)/(x - 2x^2) dx = ∫ [1/2 + 1/x + (3/2)/(1 - 2x)] dx

= 1/2 ∫ dx + ∫ 1/x dx + 3/2 ∫ 1/(1 - 2x) dx

= (1/2)x + ln|x| - (3/4)ln|1 - 2x| + C, as required.

I hope this helps!

There is no need for integration by substitution here as some answerers have done, when you have a linear argument you simply integrate as normal then divide by the derivative of the bracket.

Neither is there any need for brackets around a fraction as some answerers have, because division comes before addition by default.

Decompose the expression into partial fractions:

(1 - x²) / (x - 2x²) = (x² - 1) / (2x² - x)

(x² - 1) / (2x² - x) = (x² - 1) / x(2x - 1)

(x² - 1) / (2x² - x) = A + B / x + C / (2x - 1)

x² - 1 = A(2x² - x) + B(2x - 1) + Cx

x² - 1 = 2Ax² - Ax + 2Bx - B + Cx

x² - 1 = 2Ax² - (A - 2B - C)x - B

2A = 1

A = ½

B = 1

A - 2B - C = 0

C = A - 2B

C = ½ - 2

C = -3 / 2

(1 - x²) / (x - 2x²) = ½ + 1 / x - 3 / 2(2x - 1)

Integrate the expression term by term using this result:

∫ (1 - x²) / (x - 2x²) dx = ∫ [½ + 1 / x - 3 / 2(2x - 1)] dx

∫ (1 - x²) / (x - 2x²) dx = ∫ 1 dx / 2 + ∫ 1 / x dx - 3 ∫ 2 / (2x - 1) dx / 4

∫ (1 - x²) / (x - 2x²) dx = x / 2 + ln|x| - 3ln|2x - 1| / 4 + C

∫ (1 - x²) / (x - 2x²) dx = x / 2 - 3ln|2x - 1| / 4 + ln|x| + C

good, wen u say x2, choose thoroughly it skill x squared? next time, undergo in ideas that x squared (or any potential) might desire to be written with ^. x+a million/x=6 we could make a million/x y incredibly of a million/x. so: x+y=6 y=a million/x from the 2nd equation, all individuals comprehend: x*y=a million from the 1st equation, all individuals comprehend: x=6-y subsequently: (6-y)*y=a million 6y-y^2=a million carry this into the generally going on form: -y^2+6y-a million=0 we could element it. 2 numbers that wen extra powerful u get a million and wen u upload, u get 6. good, thats actually wat all individuals began with! any way, we could use any incorrect thank you to make certain. y=(-6+/-?(6^2-4(-a million)(-a million)))/2(-a million) y=(-6+/-?(36-4)/-2 y=(-6+/-?32)/-2 ?32 is approximately 5.66 y=(-6+/- 5.66)/-2 first fee: y=(-6+5.66)/-2 y=-.34/-2 y=.17 2nd fee: y=(-6-5.66)/-2 y=-11.66/-2 y=5.80 3 enter the 1st fee into first or 2nd equation: .17+x=6 x=6-.17 x=5.80 3 insure it suits in 2nd equation: (5.80 3)(.17)=a million u shud get .9911, a real looking answer considering that we predicted. as u can see, the 2nd fee of y is the 1st fee of x, so it is ineffective to circulate into this, by way of reality the effect could be backwards. so, x=.17 y=5.80 3 y=a million/x subsequently, x^2+a million/x^2= nonetheless approximately (.17)^2+a million/(.17)^2=.0289+a million/.0289=34.sixty 3 those have been all estimations, so do now no longer worry no keep in mind if it is a few decimals off.

I'd break it apart first:

1 / (x - 2x^2) - x^2 / (x - 2x^2) =>

1 / (x * (1 - 2x)) - x / (1 - 2x))

A/x + B / (1 - 2x) = 1 / (x * (1 - 2x))

A * (1 - 2x) + B * x = 0x + 1

A - 2Ax + Bx = 0x + 1

B - 2A = 0

A = 1

B - 2 = 0

B = 2

(1 / x + 2 / (1 - 2x) - x / (1 - 2x)) * dx

Integrate:

(1/x) * dx + 2 * dx / (1 - 2x) - x * dx / (1 - 2x)

The integral of dx/x is easy enough: ln(x)

2dx / (1 - 2x)

u = 1 - 2x

du = -2dx

-du / u

Integrate: - ln(u) => -ln(1 - 2x)

Now we have:

ln(x) - ln(1 - 2x) - int(x * dx / (1 - 2x))

u = 1 - 2x

du = -2 * dx

2x = 1 - u => x = (1 - u) / 2

x = (1 - u) / 2

dx = (-1/2) * du

(1 - 2x) = u

Now we have:

x * dx / (1 - 2x) =>

(1/2) * (1 - u) * (-1/2) * du / u =>

(1/4) * (u - 1) * du / u =>

(1/4) * (u/u - 1/u) * du =>

(1/4) * (1 - 1/u) * du

Integrate:

(1/4) * u - (1/4) * ln(u) + C

u = 1 - 2x

(1/4) * (1 - 2x) - (1/4) * ln(1 - 2x) + C

ln(x) - ln(1 - 2x) - int(x * dx / (1 - 2x)) becomes

ln(x) - ln(1 - 2x) - (1/4) * (1 - 2x - ln(1 - 2x)) + C =>

ln(x) - ln(1 - 2x) - (1/4) * (1 - 2x) + (1/4) * ln(1 - 2x) + C =>

ln(x) - (3/4) * ln(1 - 2x) + (1/4) * (2x - 1) + C =>

(2x/4) + ln(x) - (3/4) * ln(1 - 2x) - (1/4) * 1 + C =>

(x/2) + ln(x) - (3/4) * ln(1 - 2x) - 1/4 + C

Since 1/4 is a constant value, we can include it with C

(x/2) + ln(x) - (3/4) * ln(1 - 2x) + C

on integration we get: (x/2) + log(x) - (3/4)log(1-2x) + c