Trouble With Integration: ∫ e^4x (e^4x - 9)^10 dx

[tlake]

∫ e^4x (e^4x - 9)¹⁰ dx

This has gotten me pretty stumped for a while on how to proceed. I've tried integrating by parts (tabularly), putting (e^4x - 9)¹⁰ in the differentiation column and e^4x in the integration column. That gave me a very long solution of:

1/4 e^4x (e^4x - 9)¹⁰ - 10/16 e^4x (e^4x - 9)⁹ + 90/64 e^4x (e^4x - 9)⁸ - 720/256 e^4x (e^4x - 9)⁷ + 5040/1024 e^4x (e^4x - 9)⁶ - 30240/4096 e^4x (e^4x - 9)⁵ + 151200/16384 e^4x (e^4x - 9)⁴ - 604800/65536 e^4x (e^4x - 9)³ + 1814400/262144 e^4x (e^4x - 9)² - 3628800/1048576 e^4x (e^4x - 9) + 3628800/4194304 e^4x

When I checked it against Mathematica's solution, it was wrong.

I then tried a couple different substitutions:

u = e^4x, du = 1/4 e^4x dx, => 4du = e^4x dx
: ∫ e^4x (e^4x - 9)¹⁰ dx then = 4 ∫(u - 9)¹⁰ du, at which point I made another substitution of w = u - 9, so u = w + 9 and du = 10 dw
: 4 ∫(u - 9)¹⁰ du = 40∫w¹⁰ dw = 40/11 w¹¹ + C = 40/11 (e^4x - 9)¹¹ + C, again incorrect vs Mathematica.

Or what about u = e^4x - 9, du = 4e^4x - 9 dx => 9/4 du = e^4x dx
: ∫ e^4x (e^4x - 9)¹⁰ dx = 9/4 ∫ u¹⁰ du = 9/44 u¹¹ + C = 9/44 (e^4x - 9)¹¹ + C, which, surprise, was also incorrect.

I don't know what else to do, but it's probably that I'm overlooking something simple. Perhaps I've just been doing too much calculus today, and need a break. Anyway, any help you could give me would be terribly appreciated!

 

[daon2]

Or what about u = e^4x - 9, du = 4e^4x - 9 dx => 9/4 du = e^4x dx
: ∫ e^4x (e^4x - 9)¹⁰ dx = 9/4 ∫ u¹⁰ du = 9/44 u¹¹ + C = 9/44 (e^4x - 9)¹¹ + C, which, surprise, was also incorrect.

That is the method I would choose. I didn't read through all of your attempts, but here... the bold text above is not true (Not to mention somehow you turned subtraction into division or something?).

If u = e^(4x)-9, then du = 4*e^(4x)dx. So your integral becomes

$\displaystyle \frac{1}{4}\int u^{10} du$

 

[tlake]

daon2, thanks for your help. After some sleep, I came back to the problem and found it was embarrassingly easy.

∫ e^4x (e^4x - 9)¹⁰ dx

If we substitute: u = e^4x , du = 4e^4x dx , 1/4du = e^4x dx

Thus, ∫ e^4x (e^4x - 9)¹⁰ dx = 1/4 ∫ (u - 9)¹⁰ du

Substitute again: w = u - 9 , dw = du

Thus, 1/4 ∫ (u - 9)¹⁰ du = 1/4 ∫ w¹⁰ dw

1/4 ∫ w¹⁰ dw = 1/4 (1/11 w¹¹) + C = 1/44 (e^4x - 9)¹¹ + C

Thanks again!

 

[srmichael]

daon2, thanks for your help. After some sleep, I came back to the problem and found it was embarrassingly easy.

∫ e^4x (e^4x - 9)¹⁰ dx

If we substitute: u = e^4x , du = 4e^4x dx , 1/4du = e^4x dx

Thus, ∫ e^4x (e^4x - 9)¹⁰ dx = 1/4 ∫ (u - 9)¹⁰ du

Substitute again: w = u - 9 , dw = du

Thus, 1/4 ∫ (u - 9)¹⁰ du = 1/4 ∫ w¹⁰ dw

1/4 ∫ w¹⁰ dw = 1/4 (1/11 w¹¹) + C = 1/44 (e^4x - 9)¹¹ + C

Thanks again!

You can eliminate the need for the "w" substitution by doing what daon2 said and letting u = e^4x - 9. When you take the derivative of u, the 9 becomes 0

 

[tlake]

You can eliminate the need for the "w" substitution by doing what daon2 said and letting u = e^4x - 9.

Yeah, that's a smart call, and would've saved me a step or two if I had read it a little earlier. Thanks!