# Integration of polynomial.

#### paulmdrdo

##### Active member
1. ∫(x2-4x+4)4/3
2. ∫(1+1/3x)1/2dx/x2

this is what i do for number 1.

∫(x2-2)8/3

now i'm stuck.

#### Prove It

##### Well-known member
MHB Math Helper
1. ∫(x2-4x+4)4/3
2. ∫(1+1/3x)1/2dx/x2
3. ∫(3+s)1/2(s+1)2ds

this is what i do for number 1.

∫(x2-2)8/3

now i'm stuck.

That's a good start, but notice that \displaystyle \displaystyle \begin{align*} x^2 - 4x + 4 = \left( x - 2 \right) ^2 \end{align*}, not \displaystyle \displaystyle \begin{align*} \left( x^2 - 2 \right) ^2 \end{align*}, so...

\displaystyle \displaystyle \begin{align*} \int{ \left( x^2 - 4x + 4 \right) ^{\frac{4}{3}}\,dx} &= \int{ \left[ \left( x - 2 \right) ^2 \right] ^{\frac{4}{3}} \, dx} \\ &= \int{ \left( x - 2 \right) ^{\frac{8}{3}} \, dx} \end{align*}

Now let \displaystyle \displaystyle \begin{align*} u = x - 2 \implies du = dx \end{align*} and the integral becomes \displaystyle \displaystyle \begin{align*} \int{ u^{\frac{8}{3}}\,du} \end{align*}. I'm sure you can go from here.

What have you tried with the other questions?

#### soroban

##### Well-known member
Hello, paulmdrdo!

$$\displaystyle \;\;\int \left(1+\frac{1}{3x}\right)^{\frac{1}{2}}\,\frac{dx}{x^2}$$

$$\text{Let }\,u \:=\:1 + \frac{1}{3x} \:=\:1 + \frac{1}{3}x^{-1}$$

$$\text{Then: }\,du \:=\:-\frac{1}{3}x^{-2}dx \quad\Rightarrow\quad \frac{dx}{x^2} \:=\:-3\,du$$

$$\text{Substitute: }\:\int u^{\frac{1}{2}}(-3\,du) \;=\;-3\int u^{\frac{1}{2}}\,du$$

. . . . . . . $$=\;-3\cdot\tfrac{2}{3}u^{\frac{3}{2}} + C \;=\;-2u^{\frac{3}{2}} + C$$

$$\text{Back-substitute: }\:-2\left(1 + \frac{1}{3x}\right)^{\frac{3}{2}} + C$$