How would I solve $\int \left(34^3-5t^{-3}\right)dt$? I stumbled upon this section of my textbook and I don't know where to begin. We haven't reached this section in class yet but I wanted to know how to solve stuff like this. 
 A: The answer for this problem would be $=\frac{5}{2t^2}+39304t+C$
For this section of the textbook you are working with the anti-derivative. So basically you do the opposite of what you do for finding the derivative. If the integral was for 9, your answer would be $9x+C$ ( C just being a general representation of an unknown constant ).
A: HINT:
I assume you are familiar with the power rule, which states that
$$\frac{\text{d}}{\text{d}x}(x^n)=nx^{n-1}$$
Integrating both sides and dividing by $n$, we have
$$\int{\frac{\text{d}}{\text{d}x}(x^n)}\text{d}x=\int{nx^{n-1}}\text{d}x$$
$$\implies \frac{x^n}{n}=\int{x^{n-1}}\text{d}x$$
Noting that integration distributes over addition, you can consider the problem as the integral of $34^3$ and $-5t^{-3}$. Try to figure it out from here, or try guessing polynomials that when differentiated give you the resultant polynomial. As an integrator, you're attempting to undo the work of a differentiator.
A: HINT:
$$\int\left(34^3-5t^{-3}\right)\space\text{d}t=$$
$$\int 34^3\space\text{d}t-\int 5t^{-3}\space\text{d}t=$$
$$34^3\int 1\space\text{d}t-5\int t^{-3}\space\text{d}t$$
A: Hint:
for $n\neq -1$
$$\int At^ndt=A\frac{t^{n+1}}{n+1} $$
so
$$\int 34^3t^0dt+\int5t^{-3}dt$$
