# how to solve this root integral

• Feb 18th 2009, 06:36 AM
transgalactic
how to solve this root integral
i tried this:
$\displaystyle \int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}=\int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}*\frac{1+\sqrt{x+1}}{1+\sqrt{x+1}}*\frac{1-\sqrt[3]{x+1}+(x+1)^{\frac{3}{2}}}{1-\sqrt[3]{x+1}+(x+1)^{\frac{3}{2}}}$
but when i got read of 2 roots i got another two roots which are more complicated
??
• Feb 18th 2009, 06:46 AM
Jester
Quote:

Originally Posted by transgalactic
i tried this:
$\displaystyle \int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}=\int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}*\frac{1+\sqrt{x+1}}{1+\sqrt{x+1}}*\frac{1-\sqrt[3]{x+1}+(x+1)^{\frac{3}{2}}}{1-\sqrt[3]{x+1}+(x+1)^{\frac{3}{2}}}$
but when i got read of 2 roots i got another two roots which are more complicated
??

Try $\displaystyle x = u^6-1$
• Feb 18th 2009, 09:25 AM
Soroban
Hello, transgalactic!

Quote:

$\displaystyle \int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}\,dx$
Danny has the best idea . . .

Let: $\displaystyle u^6 \:=\:x+1 \quad\Rightarrow\quad x \:=\:u^6 - 1\quad\Rightarrow\quad dx \:=\:6u^5\,du$

. . Then: .$\displaystyle \sqrt{x+1} \:=\:u^3,\;\sqrt[3]{x+1} \:=\:u^2$

Substitute: .$\displaystyle \int\frac{1-u^3}{1+u^2}(6u^5\,du) \;=\;6\int\frac{u^5 - u^8}{1+u^2}\,du$

Long division: .$\displaystyle 6\int\bigg[-u^6 + u^4 + u^3 - u^2 - u + 1 + \frac{u-1}{u^2+1}\bigg]\,du$

Can you finish it now?

• Feb 18th 2009, 09:38 AM
transgalactic
$\displaystyle \int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}dx=\int \frac{(1-u^3)6u^5}{1+u^2}du=\int \frac{6u^5-6u^8}{1+u^2}du=$
$\displaystyle \int -6u^6+6u^3-6u+\frac{6u}{1+u^2}du=..$
did i devided the polynomials correctly??
• Feb 18th 2009, 03:21 PM
javax
Quote:

Originally Posted by transgalactic
$\displaystyle \int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}dx=\int \frac{(1-u^3)6u^5}{1+u^2}du=\int \frac{6u^5-6u^8}{1+u^2}du=$
$\displaystyle \int -6u^6+6u^3-6u+\frac{6u}{1+u^2}du=..$
did i devided the polynomials correctly??

as Soroban did
first factor out -6 then you'll have:
$\displaystyle (u^8-u^5):(u^2+1)=u^6-u^4-u^3+u^2+u-1-\frac{u-1}{u^2+1}$
• Feb 19th 2009, 12:04 AM
transgalactic
i got a different remainder:
-6u^6:
$\displaystyle \frac{-6u^8+6u^5}{u^2+1} \\$
$\displaystyle -6u^8+6u^5-(-6u^8-6u^6)=6u^6+6u^5\\$
6u^4:
$\displaystyle 6u^6+6u^5-(6u^6+6u^4)=6u^5-6u^4$
6u^3:
$\displaystyle 6u^5-6u^4-(6u^5+6u^3)=-6u^4-6u^3\\$
-6u^2:
$\displaystyle -6u^4-6u^3-(-6u^4-6u^2)=-6u^3+6u^2\\$
-6u:
$\displaystyle -6u^3+6u^2-(-6u^3-6u)=6u^2+6u\\$
6\\
$\displaystyle 6u^2+6u-(6u^2+6)=6+6u$
• Feb 19th 2009, 06:09 AM
Krizalid
$\displaystyle \int{\frac{u^{5}-u^{8}}{1+u^{2}}\,du}=\int{\frac{u^{5}}{1+u^{2}}\,d u}-\int{\frac{u^{8}}{1+u^{2}}\,du}.$

Put $\displaystyle v=1+u^2$ for the first integral, and

$\displaystyle \frac{u^{8}}{1+u^{2}}=\frac{u^{8}-1+1}{1+u^{2}}=\frac{\left( u^{4}+1 \right)\left( u^{2}+1 \right)\left( u^{2}-1 \right)+1}{1+u^{2}}=\left( u^{4}+1 \right)\left( u^{2}-1 \right)\,+\,\frac{1}{1+u^{2}},$

which is very easy to integrate.