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tgh^2

Integral of tgh^2 dx

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The solution

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01tanh2(x)dx\int\limits_{0}^{1} \tanh^{2}{\left(x \right)}\, dx
Integral(tanh(x)^2, (x, 0, 1))
Detail solution
  1. Let u=tanh(x)u = \tanh{\left(x \right)}.

    Then let du=(1tanh2(x))dxdu = \left(1 - \tanh^{2}{\left(x \right)}\right) dx and substitute du- du:

    (u2u21)du\int \left(- \frac{u^{2}}{u^{2} - 1}\right)\, du

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

      u2u21du=u2u21du\int \frac{u^{2}}{u^{2} - 1}\, du = - \int \frac{u^{2}}{u^{2} - 1}\, du

      1. Rewrite the integrand:

        u2u21=112(u+1)+12(u1)\frac{u^{2}}{u^{2} - 1} = 1 - \frac{1}{2 \left(u + 1\right)} + \frac{1}{2 \left(u - 1\right)}

      2. Integrate term-by-term:

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

          1du=u\int 1\, du = u

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

          (12(u+1))du=1u+1du2\int \left(- \frac{1}{2 \left(u + 1\right)}\right)\, du = - \frac{\int \frac{1}{u + 1}\, du}{2}

          1. Let u=u+1u = u + 1.

            Then let du=dudu = du and substitute dudu:

            1udu\int \frac{1}{u}\, du

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

            Now substitute uu back in:

            log(u+1)\log{\left(u + 1 \right)}

          So, the result is: log(u+1)2- \frac{\log{\left(u + 1 \right)}}{2}

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

          12(u1)du=1u1du2\int \frac{1}{2 \left(u - 1\right)}\, du = \frac{\int \frac{1}{u - 1}\, du}{2}

          1. Let u=u1u = u - 1.

            Then let du=dudu = du and substitute dudu:

            1udu\int \frac{1}{u}\, du

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

            Now substitute uu back in:

            log(u1)\log{\left(u - 1 \right)}

          So, the result is: log(u1)2\frac{\log{\left(u - 1 \right)}}{2}

        The result is: u+log(u1)2log(u+1)2u + \frac{\log{\left(u - 1 \right)}}{2} - \frac{\log{\left(u + 1 \right)}}{2}

      So, the result is: ulog(u1)2+log(u+1)2- u - \frac{\log{\left(u - 1 \right)}}{2} + \frac{\log{\left(u + 1 \right)}}{2}

    Now substitute uu back in:

    log(tanh(x)1)2+log(tanh(x)+1)2tanh(x)- \frac{\log{\left(\tanh{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\tanh{\left(x \right)} + 1 \right)}}{2} - \tanh{\left(x \right)}

  2. Add the constant of integration:

    log(tanh(x)1)2+log(tanh(x)+1)2tanh(x)+constant- \frac{\log{\left(\tanh{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\tanh{\left(x \right)} + 1 \right)}}{2} - \tanh{\left(x \right)}+ \mathrm{constant}


The answer is:

log(tanh(x)1)2+log(tanh(x)+1)2tanh(x)+constant- \frac{\log{\left(\tanh{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\tanh{\left(x \right)} + 1 \right)}}{2} - \tanh{\left(x \right)}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                                                
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 |     2             log(1 + tanh(x))             log(-1 + tanh(x))
 | tanh (x) dx = C + ---------------- - tanh(x) - -----------------
 |                          2                             2        
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tanh2(x)dx=Clog(tanh(x)1)2+log(tanh(x)+1)2tanh(x)\int \tanh^{2}{\left(x \right)}\, dx = C - \frac{\log{\left(\tanh{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\tanh{\left(x \right)} + 1 \right)}}{2} - \tanh{\left(x \right)}
The graph
0.001.000.100.200.300.400.500.600.700.800.900.01.0
The answer [src]
1 - tanh(1)
1tanh(1)1 - \tanh{\left(1 \right)}
=
=
1 - tanh(1)
1tanh(1)1 - \tanh{\left(1 \right)}
1 - tanh(1)
Numerical answer [src]
0.238405844044235
0.238405844044235
The graph
Integral of tgh^2 dx

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