Evaluate $\int_1^\infty \frac{\ln x}{x^4}$ I evaluated the integral down to $[-\frac{\ln x}{3x^3} - \frac1{9x^3}]$. So I plug in infinity and get $-1$. because you get  $-\frac\infty\infty$ for $-\frac{\ln x}{3x^3}$. $\frac1{9x^3}$ just goes to $0$. I then plug in the $1$ and get $-\frac19$. so it would be $-1 - -\frac19 $and I got $-\frac89$ which is wrong. I know the right answer is $\frac19$ but I don't understand where the $-1$ went.
 A: You correctly evaluated the indefinite integral! :)
$$\int \frac{\ln x}{x^4}\,dx=-\frac{\ln x}{3x^3}-\int \frac{1}{-3x^4}\,dx=-\frac{\ln x}{3x^3}+\frac{1}{9x^3}=\frac{-3\ln x+1}{9x^3}$$
Now we need to find
$$\int_{1}^{\infty}\frac{\ln x}{x^4}\,dx=\frac{1}{9}-\lim_{a\to\infty}\frac{-3\ln a+1}{9a^3}$$
L'Hôpital can help us :)
$$=\frac19-\lim_{a\to\infty}\frac{-\frac{3}{a}}{27a^2}=\frac{1}{9}$$
Therefore the integral converges to $\dfrac{1}{9}$.

Your mistake is in the fact that you say that $$\lim_{x\to\infty}-\frac{\ln x}{3x^3}=-1$$
First, $-\frac{\infty}{\infty}$ is a indeterminate form, and is certainly not equal to $-1$.
Second, the limit actually has a value. $3x^3$ grows significantly faster as $x$ goes to infinity compared to $\ln x$. This makes that the limit is $0$.
A: Other answers have pinpointed the OP's error, so I'll just make a remark that's a little too long for comments:  

Even if you don't know what the correct answer is, you can tell that
  $-8/9$ is wrong, because it's negative.

That is to say, the function being integrated, $(\ln x)/x^4$, is positive on the interval $(1,\infty)$, so the integral, if it exists, must also be positive.  
Ths same logic, incidentally, applies to the calculation
$$\int_{-1}^1{1\over x^2}dx={-1\over x}{\Big|}_{-1}^1={-1\over1}-{-1\over-1}=-2$$
except that in this case what's gone wrong lies in the "if it exists" condition.
A: use thie result $\int\frac{\ln(x)}{x^4}dx=-1/3\,{\frac {\ln  \left( x \right) }{{x}^{3}}}-1/9\,{x}^{-3}$ the result will be $\frac{1}{9}$
A: The limit, as $x \rightarrow \infty$, of $\frac{\ln x}{x^3}$ is not 1, it is zero.
You can't blithely divide infinity by infinity and get 1, you have to treat it a little more carefully.  In this case, l'Hospital's rule says it should be 
$$
\frac{ \frac{1}{x}
}{x^2}
$$ which obviously goes to zero.
