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Integral of (1/x)+16*x*y^2 dx

Limits of integration:

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The graph:

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Piecewise:

The solution

You have entered [src]
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 |  |1*- + 16*x*y | dx
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01(16xy2+11x)dx\int\limits_{0}^{1} \left(16 x y^{2} + 1 \cdot \frac{1}{x}\right)\, dx
Integral(1/x + 16*x*y^2, (x, 0, 1))
Detail solution
  1. Integrate term-by-term:

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

      16xy2dx=16y2xdx\int 16 x y^{2}\, dx = 16 y^{2} \int x\, dx

      1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n1n \neq -1:

        xdx=x22\int x\, dx = \frac{x^{2}}{2}

      So, the result is: 8x2y28 x^{2} y^{2}

    1. Don't know the steps in finding this integral.

      But the integral is

      log(x)\log{\left(x \right)}

    The result is: 8x2y2+log(x)8 x^{2} y^{2} + \log{\left(x \right)}

  2. Add the constant of integration:

    8x2y2+log(x)+constant8 x^{2} y^{2} + \log{\left(x \right)}+ \mathrm{constant}


The answer is:

8x2y2+log(x)+constant8 x^{2} y^{2} + \log{\left(x \right)}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                         
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 | /  1         2\             2  2         
 | |1*- + 16*x*y | dx = C + 8*x *y  + log(x)
 | \  x          /                          
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/                                           
8x2y2+logx8\,x^2\,y^2+\log x
The answer [src]
        2
oo + 8*y 
%a{\it \%a}
=
=
        2
oo + 8*y 
8y2+8 y^{2} + \infty

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