# Thread: Integration by Partial Fractions

1. ## Integration by Partial Fractions

$\int \frac{x^3}{(x+1)^{10}} dx$

The only way i can see to solve this would be to use partial fractions, but that would take a really long time. Is there another way?

2. Originally Posted by twoteenine
$\int \frac{x^3}{(x+1)^{10}} dx$

The only way i can see to solve this would be to use partial fractions, but that would take a really long time. Is there another way?
Use the substitution: $u=x+1$

3. I tried that but then I got that du=dx and then I had...

$\int \frac{x^3}{(u)^{10}} du
$

How do you change the $x^3$ to terms of u?

Edit: Nevermind i've got it, x = u-1. That was kind of obvious haha, thanks.

4. Hello, twoteenine!

Just for fun, I integrated it by parts.

. . I must get a life!

$I \;=\;\int \frac{x^3}{(x+1)^{10}}\,dx$

. . $\begin{array}{ccccccc}u &=& x^3 && dv &=& (x+1)^{10}dx \\ du &=& 3x^2\,dx && v &=& \text{-}\frac{1}{9}(x+1)^{-9} \end{array}$

$I \;=\;-\frac{1}{9}x^3(x+1)^{-9} + \frac{1}{3}\underbrace{\int x^2(x+1)^{-9}\,dx}_{\text{Integrate by parts}}$

. . $\begin{array}{ccccccc}u &=& x^2 && dv &=& (x+1)^{-9}\,dx \\ du &=& 2x\,dx && v &=&\text{-}\frac{1}{8}(x+1)^{-8} \end{array}$

$I \;=\;-\frac{1}{9}x^3(x+1)^{-9} + \frac{1}{3}\bigg[-\frac{1}{8}x^2(x+1)^{-8} + \frac{1}{4}\int x(x+1)^{-8}\,dx\bigg]$

$I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} + \frac{1}{12}\underbrace{\int x(x+1)^{-8}\,dx}_{\text{Integrate by parts}}$

. . $\begin{array}{ccccccc}u &=& x && dv &=& (x+1)^{-8}\,dx \\ du &=& dx && v &=& \text{-}\frac{1}{7}(x+1)^{-7} \end{array}$

$I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} + \frac{1}{12}\bigg[-\frac{1}{7}x(x+1)^{-7} + \frac{1}{7}\int (x+1)^{-7}\,dx\bigg]$

$I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} - \frac{1}{84}x(x+1)^{-7} + \frac{1}{84}\int(x+1)^{-7}\,dx$

$I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} - \frac{1}{84}x(x+1)^{-7} - \frac{1}{504}(x+1)^{-6} + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Technically, we're done,
. . but most textbooks simplify the answer beyond all recognition.

Factor:

. . $I \;=\;-\frac{1}{504}(x+1)^{-9}\bigg[56x^3 + 21x^2(x+1) + 6x(x+1)^2 + (x+1)^3\bigg] + C$

. . . . $=\; -\frac{1}{504}(x+1)^{-9}\left[84x^3 + 36x^2 + 9x + 1\right] + C$

. . . . $=\;-\,\frac{84x^3 + 36x^2 + 9x + 1}{504(x+1)^9} + C$

5. Originally Posted by Soroban
Hello, twoteenine!

Just for fun, I integrated it by parts.
. . I must get a life!

. . $\begin{array}{ccccccc}u &=& x^3 && dv &=& (x+1)^{10}dx \\ du &=& 3x^2\,dx && v &=& \text{-}\frac{1}{9}(x+1)^{-9} \end{array}$

$I \;=\;-\frac{1}{9}x^3(x+1)^{-9} + \frac{1}{3}\underbrace{\int x^2(x+1)^{-9}\,dx}_{\text{Integrate by parts}}$

. . $\begin{array}{ccccccc}u &=& x^2 && dv &=& (x+1)^{-9}\,dx \\ du &=& 2x\,dx && v &=&\text{-}\frac{1}{8}(x+1)^{-8} \end{array}$

$I \;=\;-\frac{1}{9}x^3(x+1)^{-9} + \frac{1}{3}\bigg[-\frac{1}{8}x^2(x+1)^{-8} + \frac{1}{4}\int x(x+1)^{-8}\,dx\bigg]$

$I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} + \frac{1}{12}\underbrace{\int x(x+1)^{-8}\,dx}_{\text{Integrate by parts}}$

. . $\begin{array}{ccccccc}u &=& x && dv &=& (x+1)^{-8}\,dx \\ du &=& dx && v &=& \text{-}\frac{1}{7}(x+1)^{-7} \end{array}$

$I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} + \frac{1}{12}\bigg[-\frac{1}{7}x(x+1)^{-7} + \frac{1}{7}\int (x+1)^{-7}\,dx\bigg]$

$I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} - \frac{1}{84}x(x+1)^{-7} + \frac{1}{84}\int(x+1)^{-7}\,dx$

$I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} - \frac{1}{84}x(x+1)^{-7} - \frac{1}{504}(x+1)^{-6} + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Technically, we're done,
. . but most textbooks simplify the answer beyond all recognition.

Factor:

. . $I \;=\;-\frac{1}{504}(x+1)^{-9}\bigg[56x^3 + 21x^2(x+1) + 6x(x+1)^2 + (x+1)^3\bigg] + C$

. . . . $=\; -\frac{1}{504}(x+1)^{-9}\left[84x^3 + 36x^2 + 9x + 1\right] + C$

. . . . $=\;-\,\frac{84x^3 + 36x^2 + 9x + 1}{504(x+1)^9} + C$

Your first v is wrong.

Edit: I mean the dv.