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Integrate integral1^e x x dx using integration by parts.

Mathematics

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Answer

e2+14\frac{e^2+1}{4}

Step-by-step solution

6 steps

To integrate ∫1exln⁡x dx\int_{1}^{e} x \ln x \, dx, we use integration by parts, which states ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du.

  1. Choose uu and dvdv.

    Let u=ln⁡xu = \ln x and dv=x dxdv = x \, dx.

  2. Find dudu and vv.

    Differentiate uu to find dudu:

    du=1x dxdu = \frac{1}{x} \, dx

    Integrate dvdv to find vv:

    v=∫x dx=x22v = \int x \, dx = \frac{x^2}{2}

  3. Apply the integration by parts formula.

    ∫xln⁡x dx=(ln⁡x)(x22)−∫(x22)(1x) dx\int x \ln x \, dx = (\ln x) \left(\frac{x^2}{2}\right) - \int \left(\frac{x^2}{2}\right) \left(\frac{1}{x}\right) \, dx ∫xln⁡x dx=x22ln⁡x−∫x2 dx\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x}{2} \, dx

  4. Evaluate the remaining integral.

    ∫x2 dx=12∫x dx=12(x22)=x24\int \frac{x}{2} \, dx = \frac{1}{2} \int x \, dx = \frac{1}{2} \left(\frac{x^2}{2}\right) = \frac{x^2}{4}

  5. Combine the results for the indefinite integral.

    ∫xln⁡x dx=x22ln⁡x−x24\int x \ln x \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4}

  6. Evaluate the definite integral from 1 to ee.

    ∫1exln⁡x dx=[x22ln⁡x−x24]1e\int_{1}^{e} x \ln x \, dx = \left[ \frac{x^2}{2} \ln x - \frac{x^2}{4} \right]_{1}^{e}

    Substitute the upper limit x=ex=e:

    (e22ln⁡e−e24)=(e22(1)−e24)=e22−e24=2e2−e24=e24\left( \frac{e^2}{2} \ln e - \frac{e^2}{4} \right) = \left( \frac{e^2}{2}(1) - \frac{e^2}{4} \right) = \frac{e^2}{2} - \frac{e^2}{4} = \frac{2e^2 - e^2}{4} = \frac{e^2}{4}

    Substitute the lower limit x=1x=1:

    (122ln⁡1−124)=(12(0)−14)=0−14=−14\left( \frac{1^2}{2} \ln 1 - \frac{1^2}{4} \right) = \left( \frac{1}{2}(0) - \frac{1}{4} \right) = 0 - \frac{1}{4} = -\frac{1}{4}

    Subtract the lower limit result from the upper limit result:

    e24−(−14)=e24+14=e2+14\frac{e^2}{4} - \left(-\frac{1}{4}\right) = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2+1}{4}

    The final answer is e2+14\frac{e^2+1}{4}.

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