<s>[INST] Find $F(x)=\int_0^xf(t)\,{\rm d}t$ where $f(x)= 1/(x-5)^2 + x^3$ Given the function $f(x)= 1/(x-5)^2 + x^3$, find $F(x)=\int_0^xf(t)\,{\rm d}t$. I'm not sure how to go about this problem since my function is in terms of $x$ and not $t$.
 [/INST] Well, $f(t)= \frac{1}{(t-5)^2} + t^3.$ 
Thus, $F(x)=\displaystyle \int_\limits 0^x \left(\frac{1}{(t-5)^2} + t^3\right)\,{\rm d}t=[-(t-5)^{-1}+\frac{t^4}{4}]_0^x=-\frac{1}{x-5}+\frac{x^4}{4}-[-\frac{1}{0-5}+\frac{0^4}{4}]=-\frac{1}{x-5}+\frac{x^4}{4}+\frac{1}{5}$
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