Mister Exam

Other calculators

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

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

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  1         
  /         
 |          
 |  1 + x   
 |  ----- dx
 |  2 - x   
 |          
/           
0
∫01x+12−x dx\int\limits_{0}^{1} \frac{x + 1}{2 - x}\, dx 
Integral((1 + x)/(2 - x), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

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

      Then let du=dxdu = dx and substitute −du- du:

      ∫(−uu−3) du\int \left(- \frac{u}{u - 3}\right)\, du

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

        ∫uu−3 du=−∫uu−3 du\int \frac{u}{u - 3}\, du = - \int \frac{u}{u - 3}\, du

        1. Rewrite the integrand:

          uu−3=1+3u−3\frac{u}{u - 3} = 1 + \frac{3}{u - 3}

        2. Integrate term-by-term:

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

            ∫1 du=u\int 1\, du = u

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

            ∫3u−3 du=3∫1u−3 du\int \frac{3}{u - 3}\, du = 3 \int \frac{1}{u - 3}\, du

            1. Let u=u−3u = u - 3.

              Then let du=dudu = du 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⁡(u−3)\log{\left(u - 3 \right)}

            So, the result is: 3log⁡(u−3)3 \log{\left(u - 3 \right)}

          The result is: u+3log⁡(u−3)u + 3 \log{\left(u - 3 \right)}

        So, the result is: −u−3log⁡(u−3)- u - 3 \log{\left(u - 3 \right)}

      Now substitute uu back in:

      −x−3log⁡(x−2)−1- x - 3 \log{\left(x - 2 \right)} - 1

    Method #2

    1. Rewrite the integrand:

      x+12−x=−1−3x−2\frac{x + 1}{2 - x} = -1 - \frac{3}{x - 2}

    2. Integrate term-by-term:

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

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

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

        ∫(−3x−2) dx=−3∫1x−2 dx\int \left(- \frac{3}{x - 2}\right)\, dx = - 3 \int \frac{1}{x - 2}\, dx

        1. Let u=x−2u = x - 2.

          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−2)\log{\left(x - 2 \right)}

        So, the result is: −3log⁡(x−2)- 3 \log{\left(x - 2 \right)}

      The result is: −x−3log⁡(x−2)- x - 3 \log{\left(x - 2 \right)}

    Method #3

    1. Rewrite the integrand:

      x+12−x=−x+1x−2\frac{x + 1}{2 - x} = - \frac{x + 1}{x - 2}

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

      ∫(−x+1x−2) dx=−∫x+1x−2 dx\int \left(- \frac{x + 1}{x - 2}\right)\, dx = - \int \frac{x + 1}{x - 2}\, dx

      1. Let u=x−2u = x - 2.

        Then let du=dxdu = dx and substitute dudu:

        ∫u+3u du\int \frac{u + 3}{u}\, du

        1. Rewrite the integrand:

          u+3u=1+3u\frac{u + 3}{u} = 1 + \frac{3}{u}

        2. Integrate term-by-term:

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

            ∫1 du=u\int 1\, du = u

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

            ∫3u du=3∫1u du\int \frac{3}{u}\, du = 3 \int \frac{1}{u}\, du

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

            So, the result is: 3log⁡(u)3 \log{\left(u \right)}

          The result is: u+3log⁡(u)u + 3 \log{\left(u \right)}

        Now substitute uu back in:

        x+3log⁡(x−2)−2x + 3 \log{\left(x - 2 \right)} - 2

      So, the result is: −x−3log⁡(x−2)+2- x - 3 \log{\left(x - 2 \right)} + 2

    Method #4

    1. Rewrite the integrand:

      x+12−x=x2−x+12−x\frac{x + 1}{2 - x} = \frac{x}{2 - x} + \frac{1}{2 - x}

    2. Integrate term-by-term:

      1. Rewrite the integrand:

        x2−x=−1−2x−2\frac{x}{2 - x} = -1 - \frac{2}{x - 2}

      2. Integrate term-by-term:

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

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

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

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

          1. Let u=x−2u = x - 2.

            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−2)\log{\left(x - 2 \right)}

          So, the result is: −2log⁡(x−2)- 2 \log{\left(x - 2 \right)}

        The result is: −x−2log⁡(x−2)- x - 2 \log{\left(x - 2 \right)}

      1. There are multiple ways to do this integral.

        Method #1

        1. Let u=2−xu = 2 - x.

          Then let du=−dxdu = - dx and substitute −du- du:

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

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

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

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

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

          Now substitute uu back in:

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

        Method #2

        1. Rewrite the integrand:

          12−x=−1x−2\frac{1}{2 - x} = - \frac{1}{x - 2}

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

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

          1. Let u=x−2u = x - 2.

            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−2)\log{\left(x - 2 \right)}

          So, the result is: −log⁡(x−2)- \log{\left(x - 2 \right)}

        Method #3

        1. Rewrite the integrand:

          12−x=−1x−2\frac{1}{2 - x} = - \frac{1}{x - 2}

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

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

          1. Let u=x−2u = x - 2.

            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−2)\log{\left(x - 2 \right)}

          So, the result is: −log⁡(x−2)- \log{\left(x - 2 \right)}

      The result is: −x−log⁡(2−x)−2log⁡(x−2)- x - \log{\left(2 - x \right)} - 2 \log{\left(x - 2 \right)}

  2. Add the constant of integration:

    −x−3log⁡(x−2)−1+constant- x - 3 \log{\left(x - 2 \right)} - 1+ \mathrm{constant}

The answer is:

−x−3log⁡(x−2)−1+constant- x - 3 \log{\left(x - 2 \right)} - 1+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                     
 |                                      
 | 1 + x                                
 | ----- dx = -1 + C - x - 3*log(-2 + x)
 | 2 - x                                
 |                                      
/
∫x+12−x dx=C−x−3log⁡(x−2)−1\int \frac{x + 1}{2 - x}\, dx = C - x - 3 \log{\left(x - 2 \right)} - 1
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.9004
Plot the graph f(x)
The answer [src] 
-1 + 3*log(2)
−1+3log⁡(2)-1 + 3 \log{\left(2 \right)} 
=
= 
-1 + 3*log(2)
−1+3log⁡(2)-1 + 3 \log{\left(2 \right)} 
-1 + 3*log(2)
Numerical answer [src] 
1.07944154167984
1.07944154167984

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

    © Mister Exam – Calculator