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Integration of polynomial.

paulmdrdo

Active member
May 13, 2013
386
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

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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 \(\displaystyle \displaystyle \begin{align*} x^2 - 4x + 4 = \left( x - 2 \right) ^2 \end{align*}\), 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
Feb 2, 2012
409
Hello, paulmdrdo!

[tex]\displaystyle [2]\;\;\int \left(1+\frac{1}{3x}\right)^{\frac{1}{2}}\,\frac{dx}{x^2}[/tex]

[tex]\text{Let }\,u \:=\:1 + \frac{1}{3x} \:=\:1 + \frac{1}{3}x^{-1}[/tex]

[tex]\text{Then: }\,du \:=\:-\frac{1}{3}x^{-2}dx \quad\Rightarrow\quad \frac{dx}{x^2} \:=\:-3\,du[/tex]

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

. . . . . . . [tex]=\;-3\cdot\tfrac{2}{3}u^{\frac{3}{2}} + C \;=\;-2u^{\frac{3}{2}} + C[/tex]

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