# 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?

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Do you know what substitution of variables means? Or the concept of an antiderivative? – Hrodelbert Jun 19 '14 at 12:00
Well I know how to integrate by substitution? But I don't think that's what you were asking. So no... Sorry? – Cookies Jun 19 '14 at 12:02
The first one is easy to integrate, the antiderivate is $\frac{(x+9)^4}{4}$. The second one is only solveable by expansion because the derivate of $x^2-9x+5$ does not appear, so substitution does not work. – Peter Jun 19 '14 at 12:06

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 $f(x)$, $$\int f'(x) \cdot (f(x))^n \,dx = \dfrac{(f(x))^{n+1}}{n+1} + C$$

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I see, that's nice to know. What if you had a power of $x$ inside. For example instead of $x$ what if it was $x^2$? – Cookies Jun 19 '14 at 12:09
What about for two functions in multiplication. For example: – Cookies Jun 19 '14 at 12:21
What about for integration of $x^2sinx$ – Cookies Jun 19 '14 at 12:22
You need to perhaps post a different question. We're all trying to chase a moving target! ;-) For the product of functions, there's the method of integration by parts, which, if you haven't learned, you will very soon! – amWhy Jun 19 '14 at 12:23

The first is simple because of the form $$\int (x+a)^n ~dx$$ hat is to say a first order polynomial to some power. So, the change of variable $x+a=y$ is clear and simple.

The second one is effectively difficult and expansion (tedious task here !) will be what I should do because I cannot find, in the most general case, a suitable change of variable for $$\int (x^2+a x+b)^n ~dx$$

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What if it is just $(x^2+5)^n$? – Cookies Jun 19 '14 at 12:17
Unfortunately, it would be almost the same ! If you change variable such as $x=y \sqrt 5$, you xill arrive to $a^{n+\frac{1}{2}} \int \cosh ^{2 n+1}(y)~dy$ wich has a nice looking but leads to a nightmare. Be sure I appreciate your curiosity in this area. Cheers :) – Claude Leibovici Jun 19 '14 at 12:32
Sorry for the typo. It is $x=\sqrt 5 \sinh(y)$ ! – Claude Leibovici Jun 19 '14 at 12:42

If $P(x)$ has degree $1$, say $P(x)=ax+b$, then $$\int P(x)^ndx=\frac1{a(n+1)}P(x)^{n+1}+C$$

If $P(x)$ has a higher degree I don't think that there is a simple formula. But any online integrator will gladly do the dirty work for you.

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Well of course it will but that's not the point. I really wanted to learn integration as the concept. – Cookies Jun 19 '14 at 12:09

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

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