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

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

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

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

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?

2. They differ by a constant C=1/6

$\displaystyle \frac{(x^{2}+1)^{3}}{6}=\frac{x^{6}}{6}+\frac{x^{4 }}{2}+\frac{x^{2}}{2}+\frac{1}{6}$

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

4. Oh, clever. Thanks guys!

5. Originally Posted by Isomorphism
Never miss the constant of integration!
If you want to know how the book did it, try the substitution $\displaystyle u = x^2$
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:
$\displaystyle \int \! x \left( u+1 \right) ^{2}\frac{du}{2}\$

or should it be:
$\displaystyle \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?

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

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

So the integral becomes

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

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

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

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

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

$\displaystyle =\frac{1}{6}(x^2+1)^3+C$

7. Originally Posted by DivideBy0

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

So the integral becomes

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

Thank you! This was the part i've totally forgot with substitution. X to the bottom Now I see the light too.

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

How it continues?

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