Answer to Question #233074 in Calculus for moe

Question #233074

Evaluate the integral "\\displaystyle{\\int_{1}^{e}(\\ln{x})^{2}dx}\u222b" using the integration by parts twice

  1. e-2
  2. e+2
  3. e
  4. 1/e
1
Expert's answer
2021-09-07T18:21:59-0400

"\\int_1^e (lnx)^2 dx\\\\\nput\\\\\nlnx=t\\implies x=e^t\\\\\ndx=e^tdt\\\\\nTherefore,\\\\\n\\int_1^e (lnx)^2 dx=\\int_0^1 t^2 e^t dt\\\\\n\\text{By using by parts method, we get}\\\\\n=t^2\\int e^tdt-\\int [\\frac{dt^2}{dt} \\int e^tdt]dt\\\\\n=t^2e^t-2\\int te^tdt\\\\\n\\text{By using by parts method, we get}\\\\\n=t^2e^t-2[te^t-e^t]\\\\\n=[t^2e^t-2te^t+2e^t]_0^1\\\\\n=e-2e+2e-2\\\\\n=e-2"

Therefore, option 1 is correct.


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