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Solving integrals/ -(ln(x))/x-1/x-c

Integral of -(ln(x))/x-1/x-c dx

Limits of integration:

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The solution

You have entered [src] 
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 |  /-log(x)    1    \   
 |  |-------- - - - c| dx
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∫01(−c+((−1)log⁡(x)x−1x)) dx\int\limits_{0}^{1} \left(- c + \left(\frac{\left(-1\right) \log{\left(x \right)}}{x} - \frac{1}{x}\right)\right)\, dx 
Integral((-log(x))/x - 1/x - c, (x, 0, 1))
Detail solution

  1. Integrate term-by-term:

    1. The integral of a constant is the constant times the variable of integration:

      ∫(−c) dx=−cx\int \left(- c\right)\, dx = - c x

    1. Integrate term-by-term:

      1. There are multiple ways to do this integral.

        Method #1

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

          Then let du=dxxdu = \frac{dx}{x} 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⁡(x)22- \frac{\log{\left(x \right)}^{2}}{2}

        Method #2

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

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

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

          1. Let u=1uu = \frac{1}{u}.

            Then let du=−duu2du = - \frac{du}{u^{2}} and substitute −du- du:

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

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

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

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

                Then let du=duudu = \frac{du}{u} 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⁡(u)22\frac{\log{\left(u \right)}^{2}}{2}

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

            Now substitute uu back in:

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

          Now substitute uu back in:

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

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

        ∫(−1x) dx=−∫1x dx\int \left(- \frac{1}{x}\right)\, dx = - \int \frac{1}{x}\, dx

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

        So, the result 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)}

    The result is: −cx−log⁡(x)22−log⁡(x)- c x - \frac{\log{\left(x \right)}^{2}}{2} - \log{\left(x \right)}

  2. Add the constant of integration:

    −cx−log⁡(x)22−log⁡(x)+constant- c x - \frac{\log{\left(x \right)}^{2}}{2} - \log{\left(x \right)}+ \mathrm{constant}

The answer is:

−cx−log⁡(x)22−log⁡(x)+constant- c x - \frac{\log{\left(x \right)}^{2}}{2} - \log{\left(x \right)}+ \mathrm{constant}

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

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

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