Evaluate the integral ∫1e(lnx)2dx∫\displaystyle{\int_{1}^{e}(\ln{x})^{2}dx}∫∫1e(lnx)2dx∫ using the integration by parts twice
∫1e(lnx)2dxputlnx=t ⟹ x=etdx=etdtTherefore,∫1e(lnx)2dx=∫01t2etdtBy using by parts method, we get=t2∫etdt−∫[dt2dt∫etdt]dt=t2et−2∫tetdtBy using by parts method, we get=t2et−2[tet−et]=[t2et−2tet+2et]01=e−2e+2e−2=e−2\int_1^e (lnx)^2 dx\\ put\\ lnx=t\implies x=e^t\\ dx=e^tdt\\ Therefore,\\ \int_1^e (lnx)^2 dx=\int_0^1 t^2 e^t dt\\ \text{By using by parts method, we get}\\ =t^2\int e^tdt-\int [\frac{dt^2}{dt} \int e^tdt]dt\\ =t^2e^t-2\int te^tdt\\ \text{By using by parts method, we get}\\ =t^2e^t-2[te^t-e^t]\\ =[t^2e^t-2te^t+2e^t]_0^1\\ =e-2e+2e-2\\ =e-2∫1e(lnx)2dxputlnx=t⟹x=etdx=etdtTherefore,∫1e(lnx)2dx=∫01t2etdtBy using by parts method, we get=t2∫etdt−∫[dtdt2∫etdt]dt=t2et−2∫tetdtBy using by parts method, we get=t2et−2[tet−et]=[t2et−2tet+2et]01=e−2e+2e−2=e−2
Therefore, option 1 is correct.
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