# Thread: Integrating an irrational function

1. ## 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.

2. 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...

3. 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.

4. 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. $(t-1)^{14}= \sum_{k=0}^{14} (-1)^{m-k}\begin{pmatrix}14 \\ k\end{pmatrix} t^k$.

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

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

Fernando Revilla

5. 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

6. Originally Posted by FernandoRevilla
I need a mathematical psychiatrist I see.

Fernando Revilla

Hahaha...

7. 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...