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Integrate ln[(x)(1x)]\ln[(x)(1-x)] in the limits 00 to 11

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Q. Integrate ln[(x)(1x)]\ln[(x)(1-x)] in the limits 00 to 11
  1. Rewrite Integral: Rewrite the integral to make it easier to work with. 01ln(x(1x))dx\int_{0}^{1} \ln(x(1-x)) \, dx
  2. Split Logarithm: Use the property of logarithms to split the lnln function.01(ln(x)+ln(1x))dx\int_{0}^{1} (\ln(x) + \ln(1-x)) \, dx
  3. Split Integrals: Split the integral into two separate integrals. 01ln(x)dx+01ln(1x)dx\int_0^1 \ln(x) \, dx + \int_0^1 \ln(1-x) \, dx
  4. Integration by Parts: Use integration by parts for ln(x)dx\int \ln(x) \, dx where u=ln(x)u=\ln(x) and dv=dxdv=dx. Let u=ln(x)u = \ln(x), then du=(1/x)dxdu = (1/x) \, dx. Let dv=dxdv = dx, then v=xv = x. Now, integrate by parts: udv=uvvdu\int u \, dv = uv - \int v \, du.
  5. Apply Integration by Parts: Apply integration by parts to the first integral. \newlinexln(x)x(1x)dxx \ln(x) - \int x \cdot (\frac{1}{x}) \, dx from 00 to 11\newlinexln(x)1dxx \ln(x) - \int 1 \, dx from 00 to 11
  6. Simplify and Integrate: Simplify and integrate. 01xln(x)xdx\int_{0}^{1} x \ln(x) - x \,dx
  7. Integration by Parts ln(1x)\ln(1-x): Now, use integration by parts for ln(1x)dx\int \ln(1-x) \, dx where u=ln(1x)u=\ln(1-x) and dv=dxdv=dx.
    Let u=ln(1x)u = \ln(1-x), then du=(11x)dxdu = -\left(\frac{1}{1-x}\right) dx.
    Let dv=dxdv = dx, then v=xv = x.
    Now, integrate by parts: udv=uvvdu\int u \, dv = uv - \int v \, du.
  8. Apply Integration by Parts: Apply integration by parts to the second integral.\newlinexln(1x)+01x(11x)dxx \ln(1-x) + \int_0^1 x \cdot \left(\frac{1}{1-x}\right) dx\newlinexln(1x)+01x1xdxx \ln(1-x) + \int_0^1 \frac{x}{1-x} dx
  9. Complex Integral ln(1x)dx\ln(1-x) \, dx: The integral x1xdx\int \frac{x}{1-x} \, dx is more complex and requires partial fractions or another method to solve.\newlineHowever, we can notice that the function ln(1x)\ln(1-x) is not defined at x=1x=1, which means the integral does not converge.

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