Evaluate the integral integrate of ((x^3-4x^2+3x+1)/(x^2+4))dx

Monday, December 10, 2012

Evaluate the integral integrate of ((x^3-4x^2+3x+1)/(x^2+4))dx

Since the
degree of numerator is larger than degree of denominator, you need to use reminder theorem such
that:

$\displaystyle{A}{\left({x}\right)}={B}{\left({x}\right)}{C}{\left({x}\right)}+{R}{\left({x}\right)}$

Notice that C(x) represents
the quotient and R(x) represents the reminder.

$\displaystyle{{x}}^{{3}}-{4}{{x}}^{{2}}+{3}{x}+{1}={\left({{x}}^{{2}}+{4}\right)}{\left({a}{x}+{b}\right)}$
+ cx + d

$\displaystyle{{x}}^{{3}}-{4}{{x}}^{{2}}+{3}{x}+{1}={a}{{x}}^{{3}}+{b}{{x}}^{{2}}+{4}{a}{x}+{4}{b}+{c}{x}+{d}$

$\displaystyle{{x}}^{{3}}-{4}{{x}}^{{2}}+{3}{x}+{1}={a}{{x}}^{{3}}+{b}{{x}}^{{2}}+{x}{\left({4}{a}+{c}\right)}+{4}{b}+{d}$

Equating the
coefficients of like powers yields:

$\displaystyle{a}={1}$

$\displaystyle{b}=$
-4

$\displaystyle{4}{a}+{c}={3}\Rightarrow{c}=-{1}$

$\displaystyle{4}{b}+{d}={1}\Rightarrow-{16}+{d}=$
1 => d = 17

$\displaystyle{{x}}^{{3}}-{4}{{x}}^{{2}}+{3}{x}+{1}={\left({{x}}^{{2}}+{4}\right)}{\left({x}-{4}\right)}-{x}+{17}$

Dividing both sides by $\displaystyle{{x}}^{{2}}+{4}$ yields:

$\displaystyle\frac{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{3}{x}+{1}}}{{{{x}}^{{2}}+{4}}}=$
x - 4 + (- x + 17)/(x^2+4)

Integrating both sides yields:

$\displaystyle\int\frac{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{3}{x}+{1}}}{{{{x}}^{{2}}+{4}}}{\left.{d}{x}\right.}=\int{x}{\left.{d}{x}\right.}-{4}\int{\left.{d}{x}\right.}+\int{\left(-{x}+\right.}$
17)/(x^2+4)dx

$\displaystyle\int\frac{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{3}{x}+{1}}}{{{{x}}^{{2}}+{4}}}{\left.{d}{x}\right.}=\frac{{{x}}^{{2}}}{{2}}-{4}{x}-\int\frac{{x}}{{{{x}}^{{2}}+{4}}}{\left.{d}{x}\right.}$
+17 int 1/(x^2+4) dx

You should use the following substitution to solve $\displaystyle\int$
x/(x^2+4) dx $\displaystyle{s}{u}{c}{h}{t}{\hat{:}}$

$\displaystyle{{x}}^{{2}}+{4}={t}\Rightarrow{2}{x}{\left.{d}{x}\right.}={\left.{d}{t}\right.}\Rightarrow{x}{\left.{d}{x}\right.}=$
(dt)/2

$\displaystyle\int\frac{{x}}{{{{x}}^{{2}}+{4}}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{2}}\right)}\int\frac{{{\left.{d}{t}\right.}}}{{t}}={\left(\frac{{1}}{{2}}\right)}{\ln}{\left|{t}\right|}+{c}$

$\displaystyle\int\frac{{{{x}}^{{3}}-{4}{{x}}^{{2}}+{3}{x}+{1}}}{{{{x}}^{{2}}+{4}}}{\left.{d}{x}\right.}=\frac{{{x}}^{{2}}}{{2}}-{4}{x}-{\left(\frac{{1}}{{2}}\right)}{\ln{{\left({{x}}^{{2}}+{4}\right)}}}+\frac{{17}}{{2}}\cdot{\arctan{{\left(\frac{{x}}{{2}}\right)}}}+$
c

Hence, evaluating the given integral yields $\displaystyle\int$
(x^3-4x^2+3x+1)/(x^2+4) dx = x^2/2 - 4x - (1/2)ln(x^2+4) + 17/2*arctan (x/2) + c.

Spooky Notes

Monday, December 10, 2012

Evaluate the integral integrate of ((x^3-4x^2+3x+1)/(x^2+4))dx

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