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,

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

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