Integrating powers of linear and quadratic functions How can I integrate function such as $(x+9)^3$? I obviously know that I can expand the function and integrate it normally. However, that is possible and feasible only as it is of third degree. What if the function is more like: $(x^2-9x+5)^7$? How could I integrate this function?
Setting $u = x+9 \implies du = \,dx$

That gives us $$\int (x+9)^3\,dx = \int u^3 \,du = \frac {u^4}4 + C = \frac 14(x+9)^4 + C$$

In your second example, we have a quadratic raised to a power, so the same method may not work. If you have $$\int x(x^2-9x+5)^7\,dx$$ then it will work since $x\,dx$ can be multiplied by a constant to give us $(x^2 - 9x + 5)' = 2x$.

With quadratics (raised to a power) for which you can complete the square, you can use trigonometric substitution. For example, $$(x^2 + 4x + 13) = (x^2 + 4x + 4 + 9) = (x+2)^2 + 3^2$$ Then the substitution $x+2 = 3 \tan \theta$ works well.

Note that for any function $f(x)$, $$\int f'(x) \cdot (f(x))^n \,dx = \dfrac{(f(x))^{n+1}}{n+1} + C$$
Nothing impossible with expansion (using the multinomial theorem):
$$(x^2-9x+5)^7=x^{14}-63x^{13}+1736x^{12}-27405x^{11}+272685x^{10}-1773954x^9+7594237x^8-21168009x^7+37971185x^6-44348850x^5+34085625x^4-17128125x^3+5425000x^2-984375x+78125$$
$$\int (x^2-9x+5)^7 dx=x^{15}/15-9x^{14}/2+1736x^{13}/13-9135x^{12}/4+272685x^{11}/11-886977x^{10}/5+7594237x^9/9-21168009x^8/8+5424455x^7-7391475x^6+6817125x^5-17128125x^4/4+5425000x^3/3-984375x^2/2+78125x+C$$