How do you integrate #int (x^3+2)dx#?

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Answer 1 · 2016-11-01T02:13:43

#int(x^3+2)dx=x^4/4+2x+c#

Using the power rule of integration, #intx^ndx=x^(n+1)/(n+1)+c,#

#int(x^3+2)dx=int(x^3+2x^0)#

#=x^(3+1)/(3+1)+(2x^1)/1+c#

#=x^4/4+2x+c#

Remember to add the constant c, since this is an indefinite integral and constants are removed when the primitive is derived.

Answer 2 · 2016-11-01T02:16:06

#intx^3+2dx#

#=(x^(3+1))/(3+1)+2x+C#

#=1/4x^4+2x+C#

This is because:

#intx^ndx#

#=(x^(n+1))/(n+1)+C#

And also:

#intkdx#

#=kx+C#