how to solve this root integral

**transgalactic** · February 18th 2009, 06:36 AM

i tried this:
$ \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
??

**Danny** · February 18th 2009, 06:46 AM

Originally Posted by transgalactic

i tried this:
$ \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 $x = u^6-1$

**Soroban** · February 18th 2009, 09:25 AM

Hello, transgalactic!

$\int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}\,dx$

Danny has the best idea . . .

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

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

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

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

Can you finish it now?

**transgalactic** · February 18th 2009, 09:38 AM

$ \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= $
$ \int -6u^6+6u^3-6u+\frac{6u}{1+u^2}du=.. $
did i devided the polynomials correctly??

**javax** · February 18th 2009, 03:21 PM

Originally Posted by transgalactic

$ \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= $
$ \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:
$(u^8-u^5)<img src=$ u^2+1)=u^6-u^4-u^3+u^2+u-1-\frac{u-1}{u^2+1}" alt="(u^8-u^5)

u^2+1)=u^6-u^4-u^3+u^2+u-1-\frac{u-1}{u^2+1}" />

**transgalactic** · February 19th 2009, 12:04 AM

i got a different remainder:
-6u^6:
$ \frac{-6u^8+6u^5}{u^2+1} \\ $
$ -6u^8+6u^5-(-6u^8-6u^6)=6u^6+6u^5\\ $
6u^4:
$ 6u^6+6u^5-(6u^6+6u^4)=6u^5-6u^4 $
6u^3:
$ 6u^5-6u^4-(6u^5+6u^3)=-6u^4-6u^3\\ $
-6u^2:
$ -6u^4-6u^3-(-6u^4-6u^2)=-6u^3+6u^2\\ $
-6u:
$ -6u^3+6u^2-(-6u^3-6u)=6u^2+6u\\ $
6\\
$ 6u^2+6u-(6u^2+6)=6+6u $

**Krizalid** · February 19th 2009, 06:09 AM

$\int{\frac{u^{5}-u^{8}}{1+u^{2}}\,du}=\int{\frac{u^{5}}{1+u^{2}}\,d u}-\int{\frac{u^{8}}{1+u^{2}}\,du}.$

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

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

which is very easy to integrate.

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