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int x ln(x+1)dx

∫xln⁡(x+1)dx \int x \ln (x+1) d x

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Full solution

Q. ∫xln⁡(x+1)dx \int x \ln (x+1) d x
  1. Choose uu and dvdv: Let's integrate by parts using the formula ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du. Choose u=ln⁡(x+1)u = \ln(x+1) and dv=x dxdv = x\,dx.
  2. Find dudu and vv: Differentiate uu to get dudu and integrate dvdv to get vv. So, du=1(x+1)dxdu = \frac{1}{(x+1)}dx and v=x22v = \frac{x^2}{2}.
  3. Apply integration by parts formula: Plug uu, dudu, vv, and dvdv into the integration by parts formula. So, ∫xln⁡(x+1)dx=(ln⁡(x+1))(x22)−∫(x22)(1(x+1))dx\int x \ln(x+1)dx = (\ln(x+1))(\frac{x^2}{2}) - \int(\frac{x^2}{2})(\frac{1}{(x+1)})dx.
  4. Simplify the integral: Simplify the integral on the right. We get ∫(x22)(1x+1)dx=∫(x2−14)dx\int(\frac{x^2}{2})(\frac{1}{x+1})dx = \int(\frac{x}{2} - \frac{1}{4})dx after dividing x2x^2 by (x+1)(x+1).

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