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integration help: integral of [x^3+8]/[2x^3+x^2+2x]
- Thread starter tacky_w
- Start date
This one ain't pretty. It will involve an Ln and an arctan.
Just glancing at it, you could try the partial fractions thing and rewrite it as:
\(\displaystyle \frac{-17x}{2(2x^{2}+x+2)}-\frac{5}{2x^{2}+x+2}+\frac{4}{x}+\frac{1}{2}\)
The two on the end are easy. Let's concentrate on the left two.
The trick is to get something in the numerator, so that when you take the derivative of your sub you have something to replace it with. You'll see what I mean when I make the subs.
By completing the square, we can rewrite as:
\(\displaystyle \frac{-17}{8}\int\frac{4x+1}{2x^{2}+x+2}dx-\frac{23}{8}\int\frac{1}{2x^{2}+x+2}dx\)
For the left one, let \(\displaystyle u=2x^{2}+2x+2, \;\ du=(4x+1)dx\)
\(\displaystyle \frac{-17}{8}\int\frac{1}{u}du\)
I will leave you finish that piece.
Now the right one:
\(\displaystyle \frac{-23}{8}\int\frac{1}{2x^{2}+x+2}dx\)
Let \(\displaystyle x=\frac{4u-1}{4}, \;\ du=dx\)
This sub gives us:
\(\displaystyle -23\int\frac{1}{16u^{2}+15}du\)
Now, an arctan sub will probably do it for you. I think letting \(\displaystyle u=\frac{\sqrt{15}}{4}tan(w), \;\ du=\frac{\sqrt{15}}{4}sec^{2}(w)dw\). When you sub in u , you get \(\displaystyle \frac{1}{15}sec^{2}(w)\). Therefore, the sec cancels and you are left with something very easy to integrate. Can you try and finish now?.
I can see why you had trouble.
Just glancing at it, you could try the partial fractions thing and rewrite it as:
\(\displaystyle \frac{-17x}{2(2x^{2}+x+2)}-\frac{5}{2x^{2}+x+2}+\frac{4}{x}+\frac{1}{2}\)
The two on the end are easy. Let's concentrate on the left two.
The trick is to get something in the numerator, so that when you take the derivative of your sub you have something to replace it with. You'll see what I mean when I make the subs.
By completing the square, we can rewrite as:
\(\displaystyle \frac{-17}{8}\int\frac{4x+1}{2x^{2}+x+2}dx-\frac{23}{8}\int\frac{1}{2x^{2}+x+2}dx\)
For the left one, let \(\displaystyle u=2x^{2}+2x+2, \;\ du=(4x+1)dx\)
\(\displaystyle \frac{-17}{8}\int\frac{1}{u}du\)
I will leave you finish that piece.
Now the right one:
\(\displaystyle \frac{-23}{8}\int\frac{1}{2x^{2}+x+2}dx\)
Let \(\displaystyle x=\frac{4u-1}{4}, \;\ du=dx\)
This sub gives us:
\(\displaystyle -23\int\frac{1}{16u^{2}+15}du\)
Now, an arctan sub will probably do it for you. I think letting \(\displaystyle u=\frac{\sqrt{15}}{4}tan(w), \;\ du=\frac{\sqrt{15}}{4}sec^{2}(w)dw\). When you sub in u , you get \(\displaystyle \frac{1}{15}sec^{2}(w)\). Therefore, the sec cancels and you are left with something very easy to integrate. Can you try and finish now?.
I can see why you had trouble.
Hello, tacky_w!
As galactus said: it ain't pretty . . .
\(\displaystyle \text{Long division: }\;\frac{x^3+8}{2x^3 + 2x^2 + 2x} \;=\;\frac{1}{2} \:- \:\frac{1}{2}\!\cdot\!\underbrace{\frac{x^2+2x-16}{x(2x^2+x+2)}}_{\text{Partial fractions}}\)
\(\displaystyle \text{We have: }\;\frac{x^2 + 2x - 16}{x(2x^2+x+2)}\;=\;\frac{A}{x} + \frac{B(4x+1)}{2x^2+x+2} + \frac{C}{2x^2+x + 2}\)
. . \(\displaystyle \text{Then: }\;x^2+2x-16 \;=\;A(2x^2+x+2) + Bx(4x+1) + Cx\)
\(\displaystyle \text{Let }x = 0\!:\;\;-16 \:=\:2 \quad\ightarrow\quad\boxed{A \:=\:-8}\)
\(\displaystyle \text{Let }x = 1\!:\;\;-13 \:=\:5A + 5B + C\quad\Rightarrow\quad 5B + C \:=\:27\;\;[1]\)
\(\displaystyle \text{Let }x = \text{-}1\!:\;\;-17 \:=\:3A + 3B - C \quad\Rightarrow\quad 3B - C \:=\:7\;\;[2]\)
\(\displaystyle \text{Add [1] and [2]: }\;8B \:=\:34\quad\Rightarrow\quad\boxed{ B\:=\:\frac{17}{4}}\)
\(\displaystyle \text{Substitute into [1]: }\;5\left(\frac{17}{4}\right) + C \:=\:27 \quad\Rightarrow\quad\boxed{ C \:=\:\frac{23}{4}}\)
\(\displaystyle \text{We have: }\;\frac{1}{2} - \frac{1}{2}\left[\frac{-8}{x} + \frac{\frac{17}{4}(4x+1)}{2x^2+x+2} + \frac{\frac{23}{4}}{2x^2+x+2}\right]\)
\(\displaystyle \text{We must integrate: }\;\frac{1}{2}\!\!\int dx \;+ \;4\!\!\int\frac{dx}{x} \;- \;\frac{17}{8}\int\!\!\frac{4x + 1}{2x^2+x+2}\,dx \;-\; \frac{23}{8}\int\!\!\frac{dx}{2x^2+x+2}\)
The first three integrals are easy.
. . The last requires completing-the-square . . .
\(\displaystyle 2x^2 +\; x \;+ \;2 \;=\;2\left(x^2 + \frac{1}{2}x + 1\right) \;=\;2\left(x^2 + \frac{1}{2}x + \frac{1}{4} + 1 - \frac{1}{4}\right) \;= \;2\bigg[\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}\bigg]\)
\(\displaystyle \text{And the last integral becomes: }\;-\frac{23}{16}\!\!\int\frac{dx}{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}\quad\hdots \text{ Arctangent!}\)
. . Got it?
As galactus said: it ain't pretty . . .
\(\displaystyle \int \frac{x^3+8}{2x^3+x^2+2x}\,dx\)
\(\displaystyle \text{Long division: }\;\frac{x^3+8}{2x^3 + 2x^2 + 2x} \;=\;\frac{1}{2} \:- \:\frac{1}{2}\!\cdot\!\underbrace{\frac{x^2+2x-16}{x(2x^2+x+2)}}_{\text{Partial fractions}}\)
\(\displaystyle \text{We have: }\;\frac{x^2 + 2x - 16}{x(2x^2+x+2)}\;=\;\frac{A}{x} + \frac{B(4x+1)}{2x^2+x+2} + \frac{C}{2x^2+x + 2}\)
. . \(\displaystyle \text{Then: }\;x^2+2x-16 \;=\;A(2x^2+x+2) + Bx(4x+1) + Cx\)
\(\displaystyle \text{Let }x = 0\!:\;\;-16 \:=\:2 \quad\ightarrow\quad\boxed{A \:=\:-8}\)
\(\displaystyle \text{Let }x = 1\!:\;\;-13 \:=\:5A + 5B + C\quad\Rightarrow\quad 5B + C \:=\:27\;\;[1]\)
\(\displaystyle \text{Let }x = \text{-}1\!:\;\;-17 \:=\:3A + 3B - C \quad\Rightarrow\quad 3B - C \:=\:7\;\;[2]\)
\(\displaystyle \text{Add [1] and [2]: }\;8B \:=\:34\quad\Rightarrow\quad\boxed{ B\:=\:\frac{17}{4}}\)
\(\displaystyle \text{Substitute into [1]: }\;5\left(\frac{17}{4}\right) + C \:=\:27 \quad\Rightarrow\quad\boxed{ C \:=\:\frac{23}{4}}\)
\(\displaystyle \text{We have: }\;\frac{1}{2} - \frac{1}{2}\left[\frac{-8}{x} + \frac{\frac{17}{4}(4x+1)}{2x^2+x+2} + \frac{\frac{23}{4}}{2x^2+x+2}\right]\)
\(\displaystyle \text{We must integrate: }\;\frac{1}{2}\!\!\int dx \;+ \;4\!\!\int\frac{dx}{x} \;- \;\frac{17}{8}\int\!\!\frac{4x + 1}{2x^2+x+2}\,dx \;-\; \frac{23}{8}\int\!\!\frac{dx}{2x^2+x+2}\)
The first three integrals are easy.
. . The last requires completing-the-square . . .
\(\displaystyle 2x^2 +\; x \;+ \;2 \;=\;2\left(x^2 + \frac{1}{2}x + 1\right) \;=\;2\left(x^2 + \frac{1}{2}x + \frac{1}{4} + 1 - \frac{1}{4}\right) \;= \;2\bigg[\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}\bigg]\)
\(\displaystyle \text{And the last integral becomes: }\;-\frac{23}{16}\!\!\int\frac{dx}{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}\quad\hdots \text{ Arctangent!}\)
. . Got it?