How do you integrate #(1/x^4) dx#?

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Answer 1 · 2018-03-23T00:14:39

#int 1/x^4color(white)(.)dx = -1/(3x^3) + C#

Note that:

#d/(dx) 1/x^3 = d/(dx) x^(-3) = -3 x^(-4) = -3(1/x^4)#

So:

#int 1/x^4color(white)(.)dx = -1/(3x^3) + C#

Answer 2 · 2018-03-23T00:16:16

#int1/x^4dx=-1/3x^-3+C#

Recall that #1/x^a=x^-a#. We can then rewrite our integral as

#intdx/x^4=intx^-4dx#

Now, recall that #intx^adx# where #ane-1# is equal to #x^(a+1)/(a+1)+C# where #C# is the constant of integration. Then,

#intx^-4dx=x^(-4+1)/(-4+1)+C=-1/3x^-3+C=-1/(3x^3)+C#