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xexp(4x^2+3)
Solving integrals/ xexp(4x^2+3)

Integral of xexp(4x^2+3) dx

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The solution

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  1               
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 |        2       
 |     4*x  + 3   
 |  x*e         dx
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0
∫01xe4x2+3 dx\int\limits_{0}^{1} x e^{4 x^{2} + 3}\, dx 
Integral(x*exp(4*x^2 + 3), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=4x2+3u = 4 x^{2} + 3.

      Then let du=8xdxdu = 8 x dx and substitute du8\frac{du}{8}:

      ∫eu8 du\int \frac{e^{u}}{8}\, du

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

        False\text{False}

        1. The integral of the exponential function is itself.

          ∫eu du=eu\int e^{u}\, du = e^{u}

        So, the result is: eu8\frac{e^{u}}{8}

      Now substitute uu back in:

      e4x2+38\frac{e^{4 x^{2} + 3}}{8}

    Method #2

    1. Rewrite the integrand:

      xe4x2+3=xe3e4x2x e^{4 x^{2} + 3} = x e^{3} e^{4 x^{2}}

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

      ∫xe3e4x2 dx=e3∫xe4x2 dx\int x e^{3} e^{4 x^{2}}\, dx = e^{3} \int x e^{4 x^{2}}\, dx

      1. Let u=4x2u = 4 x^{2}.

        Then let du=8xdxdu = 8 x dx and substitute du8\frac{du}{8}:

        ∫eu8 du\int \frac{e^{u}}{8}\, du

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

          False\text{False}

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          So, the result is: eu8\frac{e^{u}}{8}

        Now substitute uu back in:

        e4x28\frac{e^{4 x^{2}}}{8}

      So, the result is: e3e4x28\frac{e^{3} e^{4 x^{2}}}{8}

    Method #3

    1. Rewrite the integrand:

      xe4x2+3=xe3e4x2x e^{4 x^{2} + 3} = x e^{3} e^{4 x^{2}}

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

      ∫xe3e4x2 dx=e3∫xe4x2 dx\int x e^{3} e^{4 x^{2}}\, dx = e^{3} \int x e^{4 x^{2}}\, dx

      1. Let u=4x2u = 4 x^{2}.

        Then let du=8xdxdu = 8 x dx and substitute du8\frac{du}{8}:

        ∫eu8 du\int \frac{e^{u}}{8}\, du

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

          False\text{False}

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          So, the result is: eu8\frac{e^{u}}{8}

        Now substitute uu back in:

        e4x28\frac{e^{4 x^{2}}}{8}

      So, the result is: e3e4x28\frac{e^{3} e^{4 x^{2}}}{8}

  2. Now simplify:

    e4x2+38\frac{e^{4 x^{2} + 3}}{8}

  3. Add the constant of integration:

    e4x2+38+constant\frac{e^{4 x^{2} + 3}}{8}+ \mathrm{constant}

The answer is:

e4x2+38+constant\frac{e^{4 x^{2} + 3}}{8}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                              
 |                          2    
 |       2               4*x  + 3
 |    4*x  + 3          e        
 | x*e         dx = C + ---------
 |                          8    
/
∫xe4x2+3 dx=C+e4x2+38\int x e^{4 x^{2} + 3}\, dx = C + \frac{e^{4 x^{2} + 3}}{8}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.9002000
Plot the graph f(x)
The answer [src] 
   3    7
  e    e 
- -- + --
  8    8
−e38+e78- \frac{e^{3}}{8} + \frac{e^{7}}{8} 
=
= 
   3    7
  e    e 
- -- + --
  8    8
−e38+e78- \frac{e^{3}}{8} + \frac{e^{7}}{8} 
-exp(3)/8 + exp(7)/8
Numerical answer [src] 
134.568452688159
134.568452688159
The graph
Integral of xexp(4x^2+3) dx

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

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