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# Evaluate the integral:

Question:

Evaluate the integral:

$\int \frac{x-1}{3 x^{2}-4 x+3} d x$

Solution:

$I=\int \frac{x-1}{3 x^{2}-4 x+3} d x$

As we can see that there is a term of $x$ in numerator and derivative of $x^{2}$ is also $2 x$. So there is a chance that we can make substitution for $3 x^{2}-4 x+3$ and I can be reduced to a fundamental integration.

As, $\frac{\mathrm{d}}{\mathrm{dx}}\left(3 \mathrm{x}^{2}-4 \mathrm{x}+3\right)=6 \mathrm{x}-4$

$\therefore$ Let, $x-1=A(6 x-4)+B$

$\Rightarrow x-1=6 A x-4 A+B$

On comparing both sides -

We have,

$6 \mathrm{~A}=1 \Rightarrow \mathrm{A}=1 / 6$

$-4 A+B=-1 \Rightarrow B=-1+4 A=-2 / 6=-1 / 3$

Hence,

$I=\int \frac{\frac{1}{6}(6 x-4)-\frac{1}{2}}{3 x^{2}-4 x+3} d x$

$\therefore I=\frac{1}{6} \int \frac{6 x-4}{3 x^{2}-4 x+3} d x-\frac{1}{3} \int \frac{1}{3 x^{2}-4 x+3} d x$

Let, $I_{1}=\frac{1}{6} \int \frac{6 x-4}{3 x^{2}-4 x+3} d x$ and $I_{2}=\frac{1}{3} \int \frac{1}{3 x^{2}-4 x+3} d x$

Now, $|=|_{1}-\left.\right|_{2} \ldots e \operatorname{an} 1$

$\mathrm{As}, \mathrm{I}_{1}=\frac{1}{6} \int \frac{6 \mathrm{x}-4}{3 \mathrm{x}^{2}-4 \mathrm{x}+3} \mathrm{dx}$

We will solve $I_{1}$ and $I_{2}$ individually.

$\mathrm{As}, I_{1}=\frac{1}{6} \int \frac{6 \mathrm{x}-4}{3 \mathrm{x}^{2}-4 \mathrm{x}+3} \mathrm{dx}$

Let $u=3 x^{2}-4 x+3 \Rightarrow d u=(6 x-4) d x$

$\therefore I_{1}$ reduces to $\frac{1}{6} \int \frac{\mathrm{du}}{\mathrm{u}}$

Hence,

$\mathrm{I}_{1}=\frac{1}{6} \int \frac{\mathrm{du}}{\mathrm{u}}=\frac{1}{6} \log |\mathrm{u}|+\mathrm{C}\left\{\because \int \frac{\mathrm{dx}}{\mathrm{x}}=\log |\mathrm{x}|+\mathrm{C}\right\}$

On substituting value of $u$, we have:

$I_{1}=\frac{1}{6} \log \left|3 x^{2}-4 x+3\right|+C \ldots .$ eqn 2

As, $I_{2}=\frac{1}{3} \int \frac{1}{3 x^{2}-4 x+3} d x$ and we don't have any derivative of function present in denominator. $\therefore$ we will use some special integrals to solve the problem.

As denominator doesn't have any square root term. So one of the following two integrals will solve the problem.

i) $\int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C$ ii) $\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$

Now we have to reduce $\mathrm{I}_{2}$ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of $x$ is seen in the denominator.

$\therefore I_{2}=\frac{1}{9} \int \frac{1}{x^{2}-\frac{4}{3} x+1} d x$ \{on taking 3 common from denominator\}

$\Rightarrow I_{2}=\frac{1}{9} \int \frac{1}{\left(x^{2}-2\left(\frac{2}{3}\right) x+\left(\frac{2}{3}\right)^{2}\right]+1-\left(\frac{2}{3}\right)^{2}} d x$

Using: $a^{2}+2 a b+b^{2}=(a+b)^{2}$

We have:

$I_{2}=\frac{1}{9} \int \frac{1}{\left(x-\frac{2}{3}\right)^{2}+\left(\frac{\sqrt{5}}{3}\right)^{2}} d x$

$\mathrm{I}_{2}$ matches with $\int \frac{1}{\mathrm{x}^{2}+\mathrm{a}^{2}} \mathrm{dx}=\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C}$

$\therefore I_{2}=\frac{1}{9} \frac{1}{\frac{\sqrt{5}}{3}} \tan ^{-1}\left(\frac{x-\frac{2}{3}}{\frac{\sqrt{5}}{3}}\right)+C$

$\therefore \mathrm{I}_{2}=\frac{3}{9 \sqrt{5}} \tan ^{-1}\left(\frac{3 \mathrm{x}-2}{\sqrt{5}}\right)+\mathrm{C}=\frac{1}{3 \sqrt{5}} \tan ^{-1}\left(\frac{3 \mathrm{x}-2}{\sqrt{5}}\right)+\mathrm{C} \ldots$ eqn 3

From eqn 1:

$I=I_{1}-I_{2}$

Using eqn 2 and eqn 3 :

$I=\frac{1}{6} \log \left|3 x^{2}-4 x+3\right|-\frac{1}{3 \sqrt{5}} \tan ^{-1}\left(\frac{3 x-2}{\sqrt{5}}\right)+C$