Improper Integral
Calculate : $ int_0^oo x/(x^3+x^2+x+1)*dx
Risposte
Correct .
I have forgotten to calculate the value of this improper integral...
$int_0^(+oo) x/(x^3+x^2+x+1) dx = lim_(x->+oo) ( 1/2 arc tanx + 1/4 log(x^2+1) - 1/2 log(x+1) ) = pi/4
In fact, $F(x)-=1/2 arc tanx + 1/4 log(x^2+1) - 1/2 log(x+1)$
is continue in $x=0$ and $F(0)=0$, so the value of
the integral is the value of that limit.
$int_0^(+oo) x/(x^3+x^2+x+1) dx = lim_(x->+oo) ( 1/2 arc tanx + 1/4 log(x^2+1) - 1/2 log(x+1) ) = pi/4
In fact, $F(x)-=1/2 arc tanx + 1/4 log(x^2+1) - 1/2 log(x+1)$
is continue in $x=0$ and $F(0)=0$, so the value of
the integral is the value of that limit.
First of all, we see that
$x/(x^3+x^2+x+1) ~~ 1/x^2$ for $x->+oo$,
and this is the proof of the convergence of this integral.
For calculating it, it's easy to check that
$x^3 + x^2 + x +1 = (x+1)(x^2+1)$.
Then, let's search $A,B,C in RR$ so that:
$A(x^2+1)+(Bx+C)(x+1)=x$
We find, by imposing the respective relations:
$A=-1/2,B=1/2,C=1/2
So it's now very easy to calculate the integral:
$int x/(x^3+x^2+x+1) dx = 1/2 arc tanx + 1/4 log (x^2 + 1) -1/2 log(x+1)
Remember that the symbol $int f(x) dx
is often used to denote ONE primitive of $f(x)$,
therefore it's not compulsory to add the additive constant $C$.
$x/(x^3+x^2+x+1) ~~ 1/x^2$ for $x->+oo$,
and this is the proof of the convergence of this integral.
For calculating it, it's easy to check that
$x^3 + x^2 + x +1 = (x+1)(x^2+1)$.
Then, let's search $A,B,C in RR$ so that:
$A(x^2+1)+(Bx+C)(x+1)=x$
We find, by imposing the respective relations:
$A=-1/2,B=1/2,C=1/2
So it's now very easy to calculate the integral:
$int x/(x^3+x^2+x+1) dx = 1/2 arc tanx + 1/4 log (x^2 + 1) -1/2 log(x+1)
Remember that the symbol $int f(x) dx
is often used to denote ONE primitive of $f(x)$,
therefore it's not compulsory to add the additive constant $C$.
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