# Integration, which one is right? (I'm confused)

• Jan 2nd 2008, 02:49 PM
tabularasa
Integration, which one is right? (I'm confused)
I'm now little bit confused. I have textbook that says:
$\int \! x \left( {x}^{2}+1 \right) ^{2}{dx}\$
$=\frac{1}{6}\left(x ^{2}+1\right)^{3}\$

Ok, I'll do it my way and Integrate function by expand polynom first:
$\int \! x \left( {x}^{2}+1 \right) ^{2}{dx}\$
$=\mathop{\rm }\int \left(x ^{5}+2\mathop{\rm }x ^{3}+x \right)\mathop{\rm } dx$
$=\mathop{\rm }\frac{x ^{6}}{6}+\frac{x ^{4}}{2}+\frac{x ^{2}}{2}$

I get different integral function. let x=3 then first integral function gives result: 500/3=166,6666666... but seconds gives 333/2 = 166,5.

So, which one is right? (Doh)
• Jan 2nd 2008, 03:28 PM
galactus
They differ by a constant C=1/6

$\frac{(x^{2}+1)^{3}}{6}=\frac{x^{6}}{6}+\frac{x^{4 }}{2}+\frac{x^{2}}{2}+\frac{1}{6}$
• Jan 2nd 2008, 08:52 PM
Isomorphism
Quote:

Originally Posted by tabularasa
I'm now little bit confused. I have textbook that says:
$\int \! x \left( {x}^{2}+1 \right) ^{2}{dx}\$
$=\frac{1}{6}\left(x ^{2}+1\right)^{3}\$

Never miss the constant of integration!
If you want to know how the book did it, try the substitution $u = x^2$
• Jan 3rd 2008, 12:25 AM
tabularasa
Oh, clever. Thanks guys!
• Jan 3rd 2008, 02:21 AM
tabularasa
Quote:

Originally Posted by Isomorphism
Never miss the constant of integration!
If you want to know how the book did it, try the substitution $u = x^2$

Ok, I try to substitute u = x^2 Therefore I get:

u = x^2
x = u^(1/2)
du = 2x
dx = du/2

ok, then function looks like:
$
\int \! x \left( u+1 \right) ^{2}\frac{du}{2}\
$

or should it be:
$
\int \! u^{\frac{1}{2}} \left( u+1 \right) ^{2}\frac{du}{2}\
$

Could you help me little bit on this..what next or is it already wrong?
• Jan 3rd 2008, 03:17 AM
DivideBy0
Quote:

Originally Posted by tabularasa
I'm now little bit confused. I have textbook that says:
$\int \! x \left( {x}^{2}+1 \right) ^{2}{dx}\$

Let $u = x^2+1$

$\,du = 2x\,dx\Rightarrow \,dx = \frac{\,du}{2x}$

So the integral becomes

$\int x\cdot u^2 \cdot \frac{\,du}{2x}$

$=\int u^2 \cdot \frac{\,du}{2}$

$=\frac{1}{2}\int u^2 \,du$

$=\frac{1}{2} \left(\frac{u^3}{3}\right)+C$

$=\frac{1}{2}\frac{(x^2+1)^3}{3}+C$

$=\frac{1}{6}(x^2+1)^3+C$
• Jan 3rd 2008, 04:26 AM
tabularasa
Quote:

Originally Posted by DivideBy0

$\,du = 2x\,dx\Rightarrow \,dx = \frac{\,du}{2x}$

So the integral becomes

$\int x\cdot u^2 \cdot \frac{\,du}{2x}$

Thank you! This was the part i've totally forgot with substitution. X to the bottom Now I see the light too. (Rofl)
• Jan 3rd 2008, 08:12 AM
tabularasa
I have one more question.
I want to integrate function x*(x+1)^3 with substitution u = (x+1)^3

How it continues?
• Jan 3rd 2008, 08:46 AM
colby2152
Quote:

Originally Posted by tabularasa
I have one more question.
I want to integrate function x*(x+1)^3 with substitution u = (x+1)^3

How it continues?

Substitute: $u = x+1$. The integrand then becomes $(u-1)u^3$ which simplifies to $u^4 - u^3$. I think you can handle it from here.