Evaluate the integral:


Evaluate the integral:

$\int \frac{x+2}{2 x^{2}+6 x+5} d x$


$I=\int \frac{x+2}{2 x^{2}+6 x+5} d x$

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 $2 x^{2}+6 x+5$ and I can be reduced to a fundamental integration.

As, $\frac{\mathrm{d}}{\mathrm{dx}}\left(2 \mathrm{x}^{2}+6 \mathrm{x}+5\right)=4 \mathrm{x}+6$

$\therefore$ Let, $x+2=A(4 x+6)+B$

$\Rightarrow x+2=4 A x+6 A+B$

On comparing both sides -

We have,

$4 \mathrm{~A}=1 \Rightarrow \mathrm{A}=1 / 4$

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


$I=\int \frac{\frac{1}{4}(4 x+6)+\frac{1}{2}}{2 x^{2}+6 x+5} d x$

$\therefore I=\int \frac{\frac{1}{4}(4 x+6)}{2 x^{2}+6 x+5} d x+\int \frac{\frac{1}{2}}{2 x^{2}+6 x+5} d x$

Let, $I_{1}=\frac{1}{4} \int \frac{(4 x+6)}{2 x^{2}+6 x+5} d x$ and $I_{2}=\frac{1}{2} \int \frac{1}{2 x^{2}+6 x+5} d x$

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

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

As, $I_{1}=\frac{1}{4} \int \frac{(4 x+6)}{2 x^{2}+6 x+5} d x$

Let $u=2 x^{2}+6 x+5 \Rightarrow d u=(4 x+6) d x$

$\therefore \mathrm{I}_{1}$ reduces to $\frac{1}{4} \int \frac{\mathrm{du}}{\mathrm{u}}$


$I_{1}=\frac{1}{4} \int \frac{d u}{u}=\frac{1}{4} \log |u|+C\left\{\because \int \frac{d x}{x}=\log |x|+C\right\}$

On substituting value of $u$, we have:

$\mathrm{I}_{1}=\frac{1}{4} \log \left|2 \mathrm{x}^{2}+6 \mathrm{x}+5\right|+\mathrm{C} \ldots . .$ eqn 2

As, $I_{2}=\frac{1}{2} \int \frac{1}{2 x^{2}+6 x+5} d x$ 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.

$\therefore I_{2}=\frac{1}{2} \int \frac{1}{2 x^{2}+6 x+5} d x=\frac{1}{2} \int \frac{1}{2\left(x^{2}+3 x+\frac{5}{2}\right)} d x=\frac{1}{4} \int \frac{1}{x^{2}+3 x+\frac{5}{2}} d x$

$\Rightarrow I_{2}=\frac{1}{4} \int \frac{6}{\left\{x^{2}+2\left(\frac{2}{2}\right) x+\left(\frac{2}{2}\right)^{2}\right\}+\frac{5}{2}-\left(\frac{2}{2}\right)^{2}} d x$

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

We have:

$\mathrm{I}_{2}=\frac{1}{4} \int \frac{1}{\left(\mathrm{x}+\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}} \mathrm{dx}$

$\mathrm{I}_{2}$ matches with $1 \times 2+a 2 d x=1112 \tan -1 \mathrm{x}+32112+C \mathrm{l}_{2}$ matches with the form $\int \frac{1}{\mathrm{x}^{2}+\mathrm{a}^{2}} \mathrm{dx}=\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C}$

$\therefore I_{2}=\frac{1}{4} \frac{1}{\frac{1}{2}} \tan ^{-1}\left(\frac{x+\frac{3}{2}}{\frac{1}{2}}\right)+C$

$\therefore I_{2}=\frac{1}{2} \tan ^{-1}(2 x+3)+C \ldots$ eqn 3

From eqn 1, we have:


Using eqn 2 and 3 , we get -

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

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