Mister Exam

Other calculators

1/(1-x^2)^2
Solving integrals/ 1/(1-x^2)^2

Integral of 1/(1-x^2)^2 dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  1             
  /             
 |              
 |      1       
 |  --------- dx
 |          2   
 |  /     2\    
 |  \1 - x /    
 |              
/               
0
∫011(1−x2)2 dx\int\limits_{0}^{1} \frac{1}{\left(1 - x^{2}\right)^{2}}\, dx 
Integral(1/((1 - x^2)^2), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      1(1−x2)2=14(x+1)+14(x+1)2−14(x−1)+14(x−1)2\frac{1}{\left(1 - x^{2}\right)^{2}} = \frac{1}{4 \left(x + 1\right)} + \frac{1}{4 \left(x + 1\right)^{2}} - \frac{1}{4 \left(x - 1\right)} + \frac{1}{4 \left(x - 1\right)^{2}}

    2. Integrate term-by-term:

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

        ∫14(x+1) dx=∫1x+1 dx4\int \frac{1}{4 \left(x + 1\right)}\, dx = \frac{\int \frac{1}{x + 1}\, dx}{4}

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

          Then let du=dxdu = dx and substitute dudu:

          ∫1u du\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⁡(x+1)\log{\left(x + 1 \right)}

        So, the result is: log⁡(x+1)4\frac{\log{\left(x + 1 \right)}}{4}

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

        ∫14(x+1)2 dx=∫1(x+1)2 dx4\int \frac{1}{4 \left(x + 1\right)^{2}}\, dx = \frac{\int \frac{1}{\left(x + 1\right)^{2}}\, dx}{4}

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

          Then let du=dxdu = dx and substitute dudu:

          ∫1u2 du\int \frac{1}{u^{2}}\, du

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

            ∫1u2 du=−1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

          Now substitute uu back in:

          −1x+1- \frac{1}{x + 1}

        So, the result is: −14(x+1)- \frac{1}{4 \left(x + 1\right)}

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

        ∫(−14(x−1)) dx=−∫1x−1 dx4\int \left(- \frac{1}{4 \left(x - 1\right)}\right)\, dx = - \frac{\int \frac{1}{x - 1}\, dx}{4}

        1. Let u=x−1u = x - 1.

          Then let du=dxdu = dx and substitute dudu:

          ∫1u du\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⁡(x−1)\log{\left(x - 1 \right)}

        So, the result is: −log⁡(x−1)4- \frac{\log{\left(x - 1 \right)}}{4}

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

        ∫14(x−1)2 dx=∫1(x−1)2 dx4\int \frac{1}{4 \left(x - 1\right)^{2}}\, dx = \frac{\int \frac{1}{\left(x - 1\right)^{2}}\, dx}{4}

        1. Let u=x−1u = x - 1.

          Then let du=dxdu = dx and substitute dudu:

          ∫1u2 du\int \frac{1}{u^{2}}\, du

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

            ∫1u2 du=−1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

          Now substitute uu back in:

          −1x−1- \frac{1}{x - 1}

        So, the result is: −14(x−1)- \frac{1}{4 \left(x - 1\right)}

      The result is: −log⁡(x−1)4+log⁡(x+1)4−14(x+1)−14(x−1)- \frac{\log{\left(x - 1 \right)}}{4} + \frac{\log{\left(x + 1 \right)}}{4} - \frac{1}{4 \left(x + 1\right)} - \frac{1}{4 \left(x - 1\right)}

    Method #2

    1. Rewrite the integrand:

      1(1−x2)2=1x4−2x2+1\frac{1}{\left(1 - x^{2}\right)^{2}} = \frac{1}{x^{4} - 2 x^{2} + 1}

    2. Rewrite the integrand:

      1x4−2x2+1=14(x+1)+14(x+1)2−14(x−1)+14(x−1)2\frac{1}{x^{4} - 2 x^{2} + 1} = \frac{1}{4 \left(x + 1\right)} + \frac{1}{4 \left(x + 1\right)^{2}} - \frac{1}{4 \left(x - 1\right)} + \frac{1}{4 \left(x - 1\right)^{2}}

    3. Integrate term-by-term:

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

        ∫14(x+1) dx=∫1x+1 dx4\int \frac{1}{4 \left(x + 1\right)}\, dx = \frac{\int \frac{1}{x + 1}\, dx}{4}

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

          Then let du=dxdu = dx and substitute dudu:

          ∫1u du\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⁡(x+1)\log{\left(x + 1 \right)}

        So, the result is: log⁡(x+1)4\frac{\log{\left(x + 1 \right)}}{4}

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

        ∫14(x+1)2 dx=∫1(x+1)2 dx4\int \frac{1}{4 \left(x + 1\right)^{2}}\, dx = \frac{\int \frac{1}{\left(x + 1\right)^{2}}\, dx}{4}

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

          Then let du=dxdu = dx and substitute dudu:

          ∫1u2 du\int \frac{1}{u^{2}}\, du

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

            ∫1u2 du=−1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

          Now substitute uu back in:

          −1x+1- \frac{1}{x + 1}

        So, the result is: −14(x+1)- \frac{1}{4 \left(x + 1\right)}

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

        ∫(−14(x−1)) dx=−∫1x−1 dx4\int \left(- \frac{1}{4 \left(x - 1\right)}\right)\, dx = - \frac{\int \frac{1}{x - 1}\, dx}{4}

        1. Let u=x−1u = x - 1.

          Then let du=dxdu = dx and substitute dudu:

          ∫1u du\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⁡(x−1)\log{\left(x - 1 \right)}

        So, the result is: −log⁡(x−1)4- \frac{\log{\left(x - 1 \right)}}{4}

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

        ∫14(x−1)2 dx=∫1(x−1)2 dx4\int \frac{1}{4 \left(x - 1\right)^{2}}\, dx = \frac{\int \frac{1}{\left(x - 1\right)^{2}}\, dx}{4}

        1. Let u=x−1u = x - 1.

          Then let du=dxdu = dx and substitute dudu:

          ∫1u2 du\int \frac{1}{u^{2}}\, du

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

            ∫1u2 du=−1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

          Now substitute uu back in:

          −1x−1- \frac{1}{x - 1}

        So, the result is: −14(x−1)- \frac{1}{4 \left(x - 1\right)}

      The result is: −log⁡(x−1)4+log⁡(x+1)4−14(x+1)−14(x−1)- \frac{\log{\left(x - 1 \right)}}{4} + \frac{\log{\left(x + 1 \right)}}{4} - \frac{1}{4 \left(x + 1\right)} - \frac{1}{4 \left(x - 1\right)}

    Method #3

    1. Rewrite the integrand:

      1(1−x2)2=1x4−2x2+1\frac{1}{\left(1 - x^{2}\right)^{2}} = \frac{1}{x^{4} - 2 x^{2} + 1}

    2. Rewrite the integrand:

      1x4−2x2+1=14(x+1)+14(x+1)2−14(x−1)+14(x−1)2\frac{1}{x^{4} - 2 x^{2} + 1} = \frac{1}{4 \left(x + 1\right)} + \frac{1}{4 \left(x + 1\right)^{2}} - \frac{1}{4 \left(x - 1\right)} + \frac{1}{4 \left(x - 1\right)^{2}}

    3. Integrate term-by-term:

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

        ∫14(x+1) dx=∫1x+1 dx4\int \frac{1}{4 \left(x + 1\right)}\, dx = \frac{\int \frac{1}{x + 1}\, dx}{4}

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

          Then let du=dxdu = dx and substitute dudu:

          ∫1u du\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⁡(x+1)\log{\left(x + 1 \right)}

        So, the result is: log⁡(x+1)4\frac{\log{\left(x + 1 \right)}}{4}

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

        ∫14(x+1)2 dx=∫1(x+1)2 dx4\int \frac{1}{4 \left(x + 1\right)^{2}}\, dx = \frac{\int \frac{1}{\left(x + 1\right)^{2}}\, dx}{4}

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

          Then let du=dxdu = dx and substitute dudu:

          ∫1u2 du\int \frac{1}{u^{2}}\, du

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

            ∫1u2 du=−1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

          Now substitute uu back in:

          −1x+1- \frac{1}{x + 1}

        So, the result is: −14(x+1)- \frac{1}{4 \left(x + 1\right)}

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

        ∫(−14(x−1)) dx=−∫1x−1 dx4\int \left(- \frac{1}{4 \left(x - 1\right)}\right)\, dx = - \frac{\int \frac{1}{x - 1}\, dx}{4}

        1. Let u=x−1u = x - 1.

          Then let du=dxdu = dx and substitute dudu:

          ∫1u du\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⁡(x−1)\log{\left(x - 1 \right)}

        So, the result is: −log⁡(x−1)4- \frac{\log{\left(x - 1 \right)}}{4}

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

        ∫14(x−1)2 dx=∫1(x−1)2 dx4\int \frac{1}{4 \left(x - 1\right)^{2}}\, dx = \frac{\int \frac{1}{\left(x - 1\right)^{2}}\, dx}{4}

        1. Let u=x−1u = x - 1.

          Then let du=dxdu = dx and substitute dudu:

          ∫1u2 du\int \frac{1}{u^{2}}\, du

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

            ∫1u2 du=−1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

          Now substitute uu back in:

          −1x−1- \frac{1}{x - 1}

        So, the result is: −14(x−1)- \frac{1}{4 \left(x - 1\right)}

      The result is: −log⁡(x−1)4+log⁡(x+1)4−14(x+1)−14(x−1)- \frac{\log{\left(x - 1 \right)}}{4} + \frac{\log{\left(x + 1 \right)}}{4} - \frac{1}{4 \left(x + 1\right)} - \frac{1}{4 \left(x - 1\right)}

  2. Now simplify:

    −2x+(x−1)(x+1)(−log⁡(x−1)+log⁡(x+1))4(x−1)(x+1)\frac{- 2 x + \left(x - 1\right) \left(x + 1\right) \left(- \log{\left(x - 1 \right)} + \log{\left(x + 1 \right)}\right)}{4 \left(x - 1\right) \left(x + 1\right)}

  3. Add the constant of integration:

    −2x+(x−1)(x+1)(−log⁡(x−1)+log⁡(x+1))4(x−1)(x+1)+constant\frac{- 2 x + \left(x - 1\right) \left(x + 1\right) \left(- \log{\left(x - 1 \right)} + \log{\left(x + 1 \right)}\right)}{4 \left(x - 1\right) \left(x + 1\right)}+ \mathrm{constant}

The answer is:

−2x+(x−1)(x+1)(−log⁡(x−1)+log⁡(x+1))4(x−1)(x+1)+constant\frac{- 2 x + \left(x - 1\right) \left(x + 1\right) \left(- \log{\left(x - 1 \right)} + \log{\left(x + 1 \right)}\right)}{4 \left(x - 1\right) \left(x + 1\right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                    
 |                                                                     
 |     1                  1           1        log(-1 + x)   log(1 + x)
 | --------- dx = C - --------- - ---------- - ----------- + ----------
 |         2          4*(1 + x)   4*(-1 + x)        4            4     
 | /     2\                                                            
 | \1 - x /                                                            
 |                                                                     
/
∫1(1−x2)2 dx=C−log⁡(x−1)4+log⁡(x+1)4−14(x+1)−14(x−1)\int \frac{1}{\left(1 - x^{2}\right)^{2}}\, dx = C - \frac{\log{\left(x - 1 \right)}}{4} + \frac{\log{\left(x + 1 \right)}}{4} - \frac{1}{4 \left(x + 1\right)} - \frac{1}{4 \left(x - 1\right)}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90040000000
Plot the graph f(x)
The answer [src] 
     pi*I
oo + ----
      4
∞+iπ4\infty + \frac{i \pi}{4} 
=
= 
     pi*I
oo + ----
      4
∞+iπ4\infty + \frac{i \pi}{4} 
oo + pi*i/4
Numerical answer [src] 
3.45048902814162e+18
3.45048902814162e+18
The graph
Integral of 1/(1-x^2)^2 dx

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

    © Mister Exam – Calculator