How do you find the anti-derivative of #3 / x^3#?

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Answer 1 · 2018-03-20T08:09:39

#\int \frac{3}{x^3}dx#

First take out the constant "3"

#3\int \frac{1}{x^3}dx#

Apply property #\frac{1}{x^3}=x^{-3}#

#3\int \x^{-3}dx#

#3 \frac{x^{-3+1}}{-3+1}# =

#3(frac{x^{-2}}{-2})#=

#3(-\frac{1}{2x^2})#=

Final Answer:

#-\frac{3}{2x^2}+C#

Answer 2 · 2018-03-20T08:19:10

#1.5/(x^2)#

Well, the antiderivative of a function is the same thing as the integral of the function.

So, the antiderivative of #3/x^3# is the same as #int3/x^3 \ dx#.

And here we go,

#int3/x^3 \ dx#

We take the constant out, and we get,

#=3int1/x^3 \ dx#

Using a little bit of algebraic manipulation, we get

#=3int(x^3)^-1 \ dx#

#=3intx^-3 \ dx#

Now, we use the power rule, which states that #intx^n \ dx=(x^(n+1))/(n+1)#, and so we get

#=3*(x^(-3+1))/(-3+1)#

#=3*(x^-2)/-2#

#=(-3x^-2)/-2#

We also know that #x^-a=1/(x^a)#.

#=(-3)/(-2x^2)#

#=1.5/(x^2)#