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Solving integrals/ (1+log(x))/x

Integral of (1+log(x))/x dx

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The solution

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∫01log⁡(x)+1x dx\int\limits_{0}^{1} \frac{\log{\left(x \right)} + 1}{x}\, dx 
Integral((1 + log(x))/x, (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=log⁡(x)+1u = \log{\left(x \right)} + 1.

      Then let du=dxxdu = \frac{dx}{x} and substitute dudu:

      ∫u du\int u\, du

      1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

        ∫u du=u22\int u\, du = \frac{u^{2}}{2}

      Now substitute uu back in:

      (log⁡(x)+1)22\frac{\left(\log{\left(x \right)} + 1\right)^{2}}{2}

    Method #2

    1. Rewrite the integrand:

      log⁡(x)+1x=log⁡(x)x+1x\frac{\log{\left(x \right)} + 1}{x} = \frac{\log{\left(x \right)}}{x} + \frac{1}{x}

    2. Integrate term-by-term:

      1. Let u=1xu = \frac{1}{x}.

        Then let du=−dxx2du = - \frac{dx}{x^{2}} and substitute −du- du:

        ∫(−log⁡(1u)u) du\int \left(- \frac{\log{\left(\frac{1}{u} \right)}}{u}\right)\, du

        1. The integral of a constant times a function is the constant times the integral of the function:

          ∫log⁡(1u)u du=−∫log⁡(1u)u du\int \frac{\log{\left(\frac{1}{u} \right)}}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}}{u}\, du

          1. Let u=log⁡(1u)u = \log{\left(\frac{1}{u} \right)}.

            Then let du=−duudu = - \frac{du}{u} and substitute −du- du:

            ∫(−u) du\int \left(- u\right)\, du

            1. The integral of a constant times a function is the constant times the integral of the function:

              ∫u du=−∫u du\int u\, du = - \int u\, du

              1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                ∫u du=u22\int u\, du = \frac{u^{2}}{2}

              So, the result is: −u22- \frac{u^{2}}{2}

            Now substitute uu back in:

            −log⁡(1u)22- \frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}

          So, the result is: log⁡(1u)22\frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}

        Now substitute uu back in:

        log⁡(x)22\frac{\log{\left(x \right)}^{2}}{2}

      1. The integral of 1x\frac{1}{x} is log⁡(x)\log{\left(x \right)}.

      The result is: log⁡(x)22+log⁡(x)\frac{\log{\left(x \right)}^{2}}{2} + \log{\left(x \right)}

  2. Add the constant of integration:

    (log⁡(x)+1)22+constant\frac{\left(\log{\left(x \right)} + 1\right)^{2}}{2}+ \mathrm{constant}

The answer is:

(log⁡(x)+1)22+constant\frac{\left(\log{\left(x \right)} + 1\right)^{2}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                 
 |                                 2
 | 1 + log(x)          (1 + log(x)) 
 | ---------- dx = C + -------------
 |     x                     2      
 |                                  
/
∫log⁡(x)+1x dx=C+(log⁡(x)+1)22\int \frac{\log{\left(x \right)} + 1}{x}\, dx = C + \frac{\left(\log{\left(x \right)} + 1\right)^{2}}{2}
Simplify
The answer [src] 
-oo
−∞-\infty 
=
= 
-oo
−∞-\infty 
-oo
Numerical answer [src] 
-927.873417281334
-927.873417281334

    Use the examples entering the upper and lower limits of integration.

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