Evaluate the integral:

Solution:

$I=\int \frac{2 x+5}{x^{2}-x-2} d x$

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, $2 x+5=A(2 x-1)+B$

$\Rightarrow 2 x+5=2 A x-A+B$

On comparing both sides -

We have,

$2 \mathrm{~A}=2 \Rightarrow \mathrm{A}=1$

$-A+B=5 \Rightarrow B=A+5=6$

Hence,

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

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

Let, $I_{1}=\int \frac{(2 x-1)}{x^{2}-x-2} d x$ and $I_{2}=\int \frac{6}{x^{2}-x-2} d x$

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

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

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

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

$\therefore l_{1}$ reduces to $\int \frac{d u}{u}$

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:

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

As, $I_{2}=\int \frac{6}{x^{2}-x-2} d x$ 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 $I_{2}$ 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.

$\therefore I_{2}=\int \frac{6}{x^{2}-x-2} d x$

$\Rightarrow I_{2}=\int \frac{6}{\left\{x^{2}-2\left(\frac{1}{2}\right) x+\left(\frac{1}{2}\right)^{2}\right\}-2-\left(\frac{1}{2}\right)^{2}} d x$

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

We have:

$I_{2}=6 \int \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}} d x$

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

$\therefore I_{2}=\frac{6}{2\left(\frac{2}{2}\right)} \log \left|\frac{\left(x-\frac{1}{2}\right)-\frac{3}{2}}{\left(x-\frac{1}{2}\right)+\frac{2}{2}}\right|+C$

$\therefore I_{2}=\frac{6}{3} \log \left|\frac{2 x-1-3}{2 x-1+3}\right|+C=2 \log \left|\frac{2 x-4}{2 x+2}\right|+C=2 \log \left|\frac{x-2}{x+1}\right|+C \ldots e q n 3$

From eqn 1, we have:

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

Using eqn 2 and 3, we get -

$I=\log \left|x^{2}-x-2\right|+2 \log \left|\frac{x-2}{x+1}\right|+C \ldots \ldots$ ans