Evaluate $\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$ This integral
$$\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$$
is fabricated based on an interesting point which I may post later.
The question is: How would you do it?
 A: $\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx=\int_{0}^{1} \frac{(x+2)(3(x^2-x)+1)+2x-1}{3(x^2-x)+1}dx =\int_{0}^{1} (x+2+\frac{2x-1}{3(x^2-x)+1})dx=\int_{0}^{1} (x+2+\frac{(x^2-x+\frac{1}{3})'}{3(x^2-x+\frac{1}{3})})dx=[\frac{1}{2}x^2+2x+\ln(x^2-x+\frac{1}{3})]_0^{1}=\frac{5}{2}$
A: \begin{gather*}
I=\int ^{1}_{0}\left(\frac{3x^{3}}{3x( x-1) +1} +1\right) dx=\int ^{1}_{0}\frac{3x^{3} dx}{3x( x-1) +1} +1\\
Let\ I_{1} =\int ^{1}_{0}\frac{x^{3} dx}{3x( x-1) +1}\\
I_{1} =\int ^{1}_{0}\frac{( 1-x)^{3} dx}{3( 1-x)( 1-x-1) +1} =\int ^{1}_{0}\frac{( 1-x)^{3} dx}{3x( x-1) +1}\\
because\int ^{a}_{0} f( x) dx=\int ^{a}_{0} f( a-x) dx\\
Adding\ both,\\
2I_{1} =\int ^{1}_{0}\frac{x^{3} +( 1-x)^{3} dx}{3x( x-1) +1}\\
Apply\ a^{3} +b^{3} =( a+b)\left( a^{2} +b^{2} -ab\right)\\
2I_{1} =\int ^{1}_{0}\frac{3x( x-1) +1dx}{3x( x-1) +1} =1\\
I_{1} =\frac{1}{2}\\
I=3I_{1} +1=\frac{3}{2} +1=\frac{5}{2}\\
\\
\end{gather*}which I hope is the correct answer:)
A: Sligthly differently:  $$I=\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$$
$$I=3\int_{0}^{1} \frac{x^3}{3x^2-3x+1}+\int_{0}^{1} 1 dx$$
to get $$ J=\int_{0}^{1} \frac{x^3}{3x^2-3x+1}dx~~~~(1)$$
Now use $\int_{0}^{a} f(x)= \int_{0}^{1} f(a-x) dx= 3J+1$
$$J=\int_{0}^{1} \frac{(1-z)^3}{3x^2-3x+1}dx~~~(2)$$
we get $$2J=\int_{0}^{1} 1 dx \implies J=\frac{1}{2} \implies I=\frac{5}{2}$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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Direct calculation by partial fractions
\begin{eqnarray*}
\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}\,\dd x &=&\int_{0}^{1}\frac{3x^{3}+3x^{2}-3x+1}{3x^{2}-3x+1}\, \dd x\\
&=&\int_{0}^{1}\left( \frac{2x-1}{3x^{2}-3x+1}+x+2\right)\,\dd x\\
&=&\int_{0}^{1} \frac{2x-1}{3x^{2}-3x+1} \, \dd x +\int_{0}^{1} x \, \dd x+ 2\int_{0}^{1} \, \dd x\\
&\overset{u=x-\frac{1}{2}}{=}&\color{red}{\int_{-1/2}^{1/2} \frac{2(u+1/2)-1}{3(u+1/2)^{2}-3(u+1/2)+1} \, \dd u}+\int_{0}^{1} x \, \dd x+ 2\int_{0}^{1} \, \dd x\\
&=& \color{red}{0}+\frac{1}{2}+2\\
&=&\boxed{\frac{5}{2}}
\end{eqnarray*}
Note, that $\displaystyle f(u):= \frac{2(u+1/2)-1}{3(u+1/2)^{2}-3(u+1/2)+1}$ is odd function and the interval $[-1/2,1/2]$ is symmetric about $0$, so we have $\displaystyle \color{red}{\boxed{\int_{-1/2}^{1/2}f(u) \, \dd u=0}}$.
A: The integral can be written as $$\int_{0}^{1} \left(x+2+\dfrac{2x-1}{3x(1-x)+1}\right)\text dx=\dfrac 52+\int_0^1\dfrac{2x-1}{3x(1-x)+1}\text dx $$
Now, in this integral use $x \to 1-x$ and note that it will be the negative of it hence this integral will vanish leaving the answer as $\dfrac{5}2$.
A: At first, we must check whether integrand function is bounded or not. [More specifically, if it is not bounded, we must approach this problem as improper integral..] Since both numerator and denominator are continuous at [0,1], we enough to check whether denominator has 0-value or not. Since $3\left(x-\dfrac{1}{2}\right)^{2}<3x^{2}-3x+1$, we can conclude that denominator does NOT have 0 value. So if we can get indefinite integral of integrand, then we can also calculate the definite integral.
Let's manipulate integrand function so that we can get indefinite integral. At first, I want to divide numerator into denominator.
$3x^{3}+3x^{2}-3x+1$
$=3(x^{3}+1)-3+(3x^{2}-3x+1)$
$=3(x+1)(x^{2}-x+1)-3+(3x^{2}-3x+1)$
$=(x+1)(3x^{2}-3x+3)-3+(3x^{2}-3x+1)$
$=(x+1)(3x^{2}-3x+1)+2(x+1)-3+(3x^{2}-3x+1)$
$=(x+2)(3x^{2}-3x+1)+(2x-1)$
From this calculation, we can manipulate integrand as $x+2+\dfrac{2x-1}{3(x^{2}-x)+1}$.
Indefinite integral of $x+2$ is $\dfrac{x^{2}}{2}+2x$, so we only need to compute indefinite integral of the last term. To integrate it, I will first substitute $x^{2}-x$ as u, so that $u'(x)=2x-1$. So it can be changed as $\dfrac{1}{3} \dfrac{1}{u+\dfrac{1}{3}}$ and its indefinite integral is $\dfrac{1}{3} \log{\left(u+\dfrac{1}{3}\right)}$. Since u(1)=u(0) in this situation, we only need to calculate $\left[\dfrac{x^{2}}{2}+2x \right]^{1}_{0}=\dfrac{5}{2}$.
