Integrating an irrational function

Printable View

January 7th 2011, 10:43 PM
Glitch

Integrating an irrational function

The question

$\int{\frac{u^{14}}{u + 1}}$

My attempt
Normally I'd use polynomial division to produce a rational integrand, but as you can see, the large power of u causes the division process to become quite lengthy. Is there a more efficient way of producing an easily solvable integrand?

Thanks.
January 7th 2011, 11:00 PM
Prove It

If you don't want to use long-division, you can just do a trick...

$\displaystyle \frac{u^{14}}{u + 1} = \frac{u^{14} + u^{13} - u^{13}}{u + 1}$

$\displaystyle = \frac{u^{13}(u + 1)}{u + 1} - \frac{u^{13}}{u + 1}$

$\displaystyle = u^{13} - \frac{u^{13}}{u + 1}$

$\displaystyle = u^{13} - \frac{u^{13} + u^{12} - u^{12}}{u + 1}$

$\displaystyle = u^{13} - \frac{u^{12}(u + 1)}{u + 1} + \frac{u^{12}}{u + 1}$

$\displaystyle = u^{13} - u^{12} + \frac{u^{12}}{u + 1}$ etc...
January 7th 2011, 11:46 PM
FernandoRevilla

An alternative:

Using the substitution $t=u+1$ we can obtain a closed solution:

$\displaystyle\int \dfrac{u^{14}du}{u+1}=\displaystyle\int \dfrac{(t-1)^{14}dt}{t}=\displaystyle\int \dfrac{1}{t}\sum_{k=0}^{14}{(-1)^{n-k}t^k}dt}=\displaystyle\int \sum_{k=0}^{14}{(-1)^{n-k}t^{k-1}dt}=$

$\log |t|+\displaystyle\sum_{k=1}^{14}\dfrac{(-1)^{n-k}t^k}{k}+C=\log |u+1|+\displaystyle\sum_{k=1}^{14}\dfrac{(-1)^{n-k}(u+1)^k}{k}+C$

Fernando Revilla

Edited: I forgot the binomial coefficients.
January 8th 2011, 09:28 AM
HallsofIvy

Quote:

Originally Posted by FernandoRevilla

An alternative:

Using the substitution $t=u+1$ we can obtain a closed solution:

$\displaystyle\int \dfrac{u^{14}du}{u+1}=\displaystyle\int \dfrac{(t-1)^{14}dt}{t}=\displaystyle\int \dfrac{1}{t}\sum_{k=0}^{14}{(-1)^{n-k}t^k}dt}$

This is incorrect. $(t-1)^{14}$ is not equal to $\sum (-1)^{n-k}t^k$ . You forgot the binomial coefficients. $(t-1)^{14}= \sum_{k=0}^{14} (-1)^{m-k}\begin{pmatrix}14 \\ k\end{pmatrix} t^k$ .

Quote:

= $\displaystyle\int \sum_{k=0}^{14}{(-1)^{n-k}t^{k-1}dt}=$

$\log |t|+\displaystyle\sum_{k=1}^{14}\dfrac{(-1)^{n-k}t^k}{k}+C=\log |u+1|+\displaystyle\sum_{k=1}^{14}\dfrac{(-1)^{n-k}(u+1)^k}{k}+C$

Fernando Revilla
January 8th 2011, 09:36 AM
FernandoRevilla

Quote:

Originally Posted by HallsofIvy

This is incorrect. $(t-1)^{14}$ is not equal to $\sum (-1)^{n-k}t^k$ . You forgot the binomial coefficients. $(t-1)^{14}= \sum_{k=0}^{14} (-1)^{m-k}\begin{pmatrix}14 \\ k\end{pmatrix} t^k$ .

O ... ps !, of course. More than incorrect. I need a mathematical psychiatrist I see. :)

Fernando Revilla
January 8th 2011, 09:43 AM
Also sprach Zarathustra

Quote:

Originally Posted by FernandoRevilla

I need a mathematical psychiatrist I see. :)

Fernando Revilla

Hahaha...
January 8th 2011, 09:54 AM
Also sprach Zarathustra

Quote:

Originally Posted by Glitch

The question

$\int{\frac{u^{14}}{u + 1}}$

My attempt
Normally I'd use polynomial division to produce a rational integrand, but as you can see, the large power of u causes the division process to become quite lengthy. Is there a more efficient way of producing an easily solvable integrand?

Thanks.

... you forgot "du" at your integral...