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Solving integrals/ 1/(100+2t)

Integral of 1/(100+2t) dx

Limits of integration:

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Piecewise:

The solution

You have entered [src] 
  1             
  /             
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 |      1       
 |  --------- dt
 |  100 + 2*t   
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0
∫0112t+100 dt\int\limits_{0}^{1} \frac{1}{2 t + 100}\, dt 
Integral(1/(100 + 2*t), (t, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=2t+100u = 2 t + 100.

      Then let du=2dtdu = 2 dt and substitute du2\frac{du}{2}:

      ∫12u du\int \frac{1}{2 u}\, du

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

        ∫1u du=∫1u du2\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}

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

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

      Now substitute uu back in:

      log⁡(2t+100)2\frac{\log{\left(2 t + 100 \right)}}{2}

    Method #2

    1. Rewrite the integrand:

      12t+100=12(t+50)\frac{1}{2 t + 100} = \frac{1}{2 \left(t + 50\right)}

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

      ∫12(t+50) dt=∫1t+50 dt2\int \frac{1}{2 \left(t + 50\right)}\, dt = \frac{\int \frac{1}{t + 50}\, dt}{2}

      1. Let u=t+50u = t + 50.

        Then let du=dtdu = dt 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⁡(t+50)\log{\left(t + 50 \right)}

      So, the result is: log⁡(t+50)2\frac{\log{\left(t + 50 \right)}}{2}

  2. Add the constant of integration:

    log⁡(2t+100)2+constant\frac{\log{\left(2 t + 100 \right)}}{2}+ \mathrm{constant}

The answer is:

log⁡(2t+100)2+constant\frac{\log{\left(2 t + 100 \right)}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                 
 |                                  
 |     1              log(100 + 2*t)
 | --------- dt = C + --------------
 | 100 + 2*t                2       
 |                                  
/
∫12t+100 dt=C+log⁡(2t+100)2\int \frac{1}{2 t + 100}\, dt = C + \frac{\log{\left(2 t + 100 \right)}}{2}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.02.5
Plot the graph f(x)
The answer [src] 
log(102)   log(100)
-------- - --------
   2          2
−log⁡(100)2+log⁡(102)2- \frac{\log{\left(100 \right)}}{2} + \frac{\log{\left(102 \right)}}{2} 
=
= 
log(102)   log(100)
-------- - --------
   2          2
−log⁡(100)2+log⁡(102)2- \frac{\log{\left(100 \right)}}{2} + \frac{\log{\left(102 \right)}}{2} 
log(102)/2 - log(100)/2
Numerical answer [src] 
0.00990131364808986
0.00990131364808986

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

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