Evaluate the integral:


Evaluate the integral:

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


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

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 $x^{2}+2 x-4$ and I can be reduced to a fundamental integration.

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

$\therefore$ Let, $x-3=A(2 x+2)+B$

$\Rightarrow x-3=2 A x+2 A+B$

On comparing both sides -

We have,

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

$2 A+B=-3 \Rightarrow B=-3-2 A=-4$


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

$\therefore I=\frac{1}{2} \int \frac{2 \mathrm{x}+2}{\mathrm{x}^{2}+2 \mathrm{x}-4} \mathrm{dx}-4 \int \frac{1}{\mathrm{x}^{2}+2 \mathrm{x}-4} \mathrm{dx}$

Now, $I=I_{1}-4 I_{2} \ldots .$ eqn 1

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

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

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

$\therefore I_{1}$ reduces to $\frac{1}{2} \int \frac{d u}{u}$


$\mathrm{I}_{1}=\frac{1}{2} \int \frac{\mathrm{du}}{\mathrm{u}}=\frac{1}{2} \log |\mathrm{u}|+\mathrm{C}\left\{\because \int \frac{\mathrm{dx}}{\mathrm{x}}=\log |\mathrm{x}|+\mathrm{C}\right\}$

On substituting value of $u$, we have:

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

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

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.

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

$\Rightarrow I_{2}=\int \frac{1}{\left\{x^{2}+2(1) x+(1)^{2}\right]-4-(1)^{2}} d x$

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

We have:

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

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

$\therefore I_{2}=\frac{1}{2 \sqrt{5}} \log \left|\frac{x+1-\sqrt{5}}{x+1+\sqrt{5}}\right|+C \ldots$ eqn 3

From eqn 1:

$I=I_{1}-4 I_{2}$

Using eqn 2 and eqn 3:

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

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

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