Calculate the following integral:
∫ dx / x(1 - √x)^2
I bailed out after whole page of messy attempts using integration by parts.
http://integrals.wolfram.com/index.jsp?expr=1%2F%28x%281-x%5E%281%2F2%29%29%5E2%29&random=false
Wolfram's answer assumes that log x is ln(x)
hmmm. Doesn't look too bad.
u = 1-√x
x = (1-u)^2
dx = -2(1-u) du
dx/(x(1-√x)^2)
= -2(1-u) du / ((1-u)^2 * u^2)
= -2 /((1-u)*u^2) du
= -2/u^2 -2/u + 2/(u-1) du
= 2(1/u + ln(1-u) - ln(u))
Now substitute u = 1-√x to get back to x
Thanks Reiny and Steve. Now to pass the exam!
To calculate the given integral, we can make use of a substitution. Let's substitute u = √x. Then, we can express dx in terms of du as follows:
dx = 2u du
Next, we can express the integral in terms of u:
∫ dx / x(1 - √x)^2 = ∫ (2u du) / (u²(1 - u)²)
= 2 ∫ du / (u(1 - u)²)
Now, we need to decompose the integrand into partial fractions. Let's express 1 / (u(1 - u)²) as follows:
1 / (u(1 - u)²) = A / u + B / (1 - u) + C / (1 - u)²
To determine the constants A, B, and C, we equate the numerators and multiply through by the denominators. This gives us:
1 = A(1 - u)² + B(u)(1 - u) + C(u)
Expanding this equation, we get:
1 = (A - A2u + Au²) + (Bu - Bu²) + Cu
Next, we can collect terms and equate like powers of u to obtain the following equations:
A + B + C = 0 (Coefficient of u²)
-A + B + C = 0 (Coefficient of u)
A = 1 (Constant term)
From the first two equations, we can solve for B and C:
B = -1
C = 0
Now, let's rewrite our integrand using partial fractions:
∫ du / (u(1 - u)²) = ∫ A/u + B/(1 - u) + C/(1 - u)² du
= ∫ 1/u - 1/(1 - u) du
= ∫ (1/u) du - ∫ (1/(1 - u)) du
Integrating the above expressions, we have:
∫ 1/u du = ln|u| + C1
∫ 1/(1 - u) du = -ln|1 - u| + C2
Substituting back u = √x, we get:
∫ dx / x(1 - √x)^2 = 2(ln|√x| - ln|1 - √x|) + C
Simplifying further, we obtain the final result:
∫ dx / x(1 - √x)^2 = 2ln|√x / (1 - √x)| + C
And that's how you calculate the given integral!