How Do You Integrate Complex Polynomial Expressions?

[paulmdrdo1]

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.

please help!

 

[Prove It]

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.

please help!

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

[math]\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*}[/math]

Now let [math]\displaystyle \begin{align*} u = x - 2 \implies du = dx \end{align*}[/math] and the integral becomes [math]\displaystyle \begin{align*} \int{ u^{\frac{8}{3}}\,du} \end{align*}[/math]. I'm sure you can go from here.What have you tried with the other questions?

 

[soroban]

Hello, paulmdrdo!

\displaystyle [2]\;\;\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