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I=int(ln x)^(2)dx

I=(lnx)2dx I=\int(\ln x)^{2} d x

Full solution

Q. I=(lnx)2dx I=\int(\ln x)^{2} d x
  1. Recognize Problem: Recognize that the integral I=(lnx)2dxI = \int (\ln x)^2 \, dx is not a standard integral and requires integration by parts.
  2. Integration by Parts: Use integration by parts, which states that udv=uvvdu\int u \, dv = uv - \int v \, du. Choose u=(lnx)2u = (\ln x)^2 (which we will differentiate) and dv=dxdv = dx (which we will integrate).
  3. Differentiate uu: Differentiate uu to find dudu. Since u=(lnx)2u = (\ln x)^2, we use the chain rule to find du=2(lnx)(1/x)dx=2(lnx)/xdxdu = 2(\ln x)(1/x) dx = 2(\ln x)/x dx.
  4. Integrate dvdv: Integrate dvdv to find vv. Since dv=dxdv = dx, we have v=xv = x.
  5. Apply Formula: Apply the integration by parts formula: udv=uvvdu\int u \, dv = uv - \int v \, du. Substituting the chosen uu, vv, dudu, and dvdv, we get I=x(lnx)2x(2(lnx)/x)dxI = x(\ln x)^2 - \int x(2(\ln x)/x) \, dx.
  6. Simplify Integral: Simplify the integral x(2(lnx)x)dx\int x\left(\frac{2(\ln x)}{x}\right) dx to 2(lnx)dx\int 2(\ln x) dx.
  7. Recognize New Problem: Recognize that the integral 2(lnx)dx\int 2(\ln x) \, dx is another integration by parts problem. Choose u=lnxu = \ln x and dv=2dxdv = 2 \, dx.
  8. Differentiate uu: Differentiate uu to find dudu. Since u=lnxu = \ln x, we have du=(1x)dxdu = (\frac{1}{x}) dx.
  9. Integrate dvdv: Integrate dvdv to find vv. Since dv=2dxdv = 2 dx, we have v=2xv = 2x.
  10. Apply Formula: Apply the integration by parts formula again: udv=uvvdu\int u \, dv = uv - \int v \, du. Substituting the chosen uu, vv, dudu, and dvdv, we get 2(lnx)dx=2x(lnx)2xxdx\int 2(\ln x) \, dx = 2x(\ln x) - \int \frac{2x}{x} \, dx.
  11. Simplify Integral: Simplify the integral 2xxdx\int \frac{2x}{x} \, dx to 2dx\int 2 \, dx.
  12. Integrate: Integrate 2dx\int 2 \, dx to get 2x2x.
  13. Combine Parts: Combine all parts to write the final expression for the original integral II. We have I=x(lnx)2(2x(lnx)2x)+CI = x(\ln x)^2 - (2x(\ln x) - 2x) + C, where CC is the constant of integration.
  14. Final Expression: Simplify the expression to combine like terms. We have I=x(lnx)22x(lnx)+2x+CI = x(\ln x)^2 - 2x(\ln x) + 2x + C.