I'm confused. I was making up some of my own problems involving higher powers of x to integrate. For example:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\displaystyle\int x^5 e^{5x}dx[/tex]

I set about going about finding [tex]\frac{dy}{dx}[/tex] up to [tex]\frac{d^6y}{dx^6}[/tex].

[tex]u=x^5[/tex]

[tex]\frac{du}{dx}=5x^4[/tex]

[tex]\frac{d^2u}{dx^2}=20x^3[/tex]

[tex]\frac{d^3u}{dx^3}=60x^2[/tex]

[tex]\frac{d^4u}{dx^4}=120x[/tex]

[tex]\frac{d^5u}{dx^5}=120[/tex]

[tex]\frac{d^6u}{dx^6}=0[/tex]

Conversely I found [tex]v=\displaystyle\int \frac{dv}{dx}=\displaystyle\int e^{5x}dx=\frac{1}{5}e^{5x} + C_1[/tex] up to

[tex]\displaystyle\int \displaystyle\int \displaystyle\int \displaystyle\int \displaystyle\int e^{5x}dx=\frac{1}{3125}e^{5x} + C_5[/tex]

I'm not even sure if you can write 5 integrals next to eachother. If not, what's the correct notation?

Anyway, I then substituted the above values into the IBP formula:

[tex]\displaystyle\int u\frac{dv}{dx}dx=uv - \displaystyle\intv\frac{du}{dx}dx[/tex]

[tex]\displaystyle\int x^5e^{5x}dx=\frac{1}{5}x^5e^5x-[\displaystyle\int x^4e^{5x}dx][/tex]

[tex]\frac{1}{5}x^5e^{5x}=[\displaystyle\int \frac{5}{4}x^4e^{5x}dx]=\frac{5}{25}x^4e^{5x} - \displaystyle\int \frac{20}{25}x^3e^{5x}dx[/tex]

[tex]\frac{1}{5}x^5e^{5x}-\frac{5}{25}x^4e^{5x}= [\displaystyle\int \frac{20}{25}x^3e^{5x}dx]=\frac{20}{125}x^3e^{5x}- [\displaystyle\int \frac{60}{125}x^2e^{5x}dx][/tex]... You get the idea.

It's kind of like reduction, reducing the integral to obtain an answer if that makes sense?

Anyway, if there's an easier way to integrate higher powers of x using IBP I'd like to know.

There's a pattern there but I can't uncover it. Sort of like the Binomial expansion, same principle it seems.

Thanks! :)

My working will help too.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Integration by parts - Does this make sense?

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