Evaluate the integral:

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

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 $-x^{2}+x+2$ and I can be reduced to a fundamental integration.

As, $\frac{\mathrm{d}}{\mathrm{dx}}\left(-\mathrm{x}^{2}+\mathrm{x}+2\right)=-2 \mathrm{x}+1$

$\therefore$ Let, $2 x=A(-2 x+1)+B$

$\Rightarrow 2 x=-2 A x+A+B$

On comparing both sides -

We have,

$-2 A=2 \Rightarrow A=-1$

$A+B=0 \Rightarrow B=-A=1$

Hence,

$\mathrm{I}=\int \frac{-(-2 \mathrm{x}+1)+1}{2+\mathrm{x}-\mathrm{x}^{2}} \mathrm{dx}$

$\therefore \mathrm{I}=-\int \frac{(-2 \mathrm{x}+1)}{2+\mathrm{x}-\mathrm{x}^{2}} \mathrm{dx}+\int \frac{1}{2+\mathrm{x}-\mathrm{x}^{2}} \mathrm{dx}$

Let, $I_{1}=-\int \frac{(-2 x+1)}{2+x-x^{2}} d x$ and $I_{2}=\int \frac{1}{2+x-x^{2}} d x$

Now, $I=I_{1}+I_{2} \ldots$ eqn $I$

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

$\mathrm{As}, \mathrm{I}_{1}=-\int \frac{(-2 \mathrm{x}+1)}{2+\mathrm{x}-\mathrm{x}^{2}} \mathrm{dx}$

Let $u=2+x-x^{2} \Rightarrow d u=(-2 x+1) d x$

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

Hence,

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

On substituting value of $u$, we have:

$\mathrm{I}_{1}=-\log \left|2+\mathrm{x}-\mathrm{x}^{2}\right|+\mathrm{C} \ldots .$ eqn 2

As, $I_{2}=\int \frac{1}{2+x-x^{2}} d x$ 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 denominator.

$\therefore I_{2}=-\int \frac{1}{x^{2}-x-2} d x$

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

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

We have:

$\mathrm{I}_{2}=-\int \frac{1}{\left(\mathrm{x}-\frac{1}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}} \mathrm{dx}$

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

$\therefore I_{2}=-\frac{1}{2\left(\frac{2}{2}\right)} \log \left|\frac{\left(x-\frac{1}{2}\right)-\frac{3}{2}}{\left(x-\frac{1}{2}\right)+\frac{2}{2}}\right|+C$

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

From eqn 1:

$\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}$

Using eqn 2 and eqn 3 :

$\therefore|=-\log | 2+\mathrm{x}-\mathrm{x}^{2}\left|-\frac{1}{3} \log \right| \frac{\mathrm{x}-2}{\mathrm{x}+1} \mid+\mathrm{C}$