# Evaluate the integral:

Question:

Evaluate the integral:

$\int \frac{x+7}{3 x^{2}+25 x+28} d x$

Solution:

$I=\int \frac{x+7}{3 x^{2}+25 x+28} d x$

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 $3 x^{2}+13 x-10$ and I can be reduced to a fundamental integration.

As, $\frac{\mathrm{d}}{\mathrm{dx}}\left(3 \mathrm{x}^{2}+25 \mathrm{x}+28\right)=6 \mathrm{x}+25$

$\therefore$ Let, $x+7=A(6 x+25)+B$

$\Rightarrow x+7=6 A x+25 A+B$

On comparing both sides -

We have,

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

$25 A+B=5 \Rightarrow B=-25 A+5=5 / 6$

Hence,

$I=\int \frac{\frac{1}{6}(6 x+25)+\frac{5}{6}}{3 x^{2}+25 x+28} d x$

$\therefore I=\int \frac{\frac{1}{6}(6 x+25)}{3 x^{2}+25 x+28} d x+\int \frac{\frac{5}{6}}{3 x^{2}+25 x+28} d x$

Let, $I_{1}=\frac{1}{6} \int \frac{(6 x+25)}{3 x^{2}+25 x+28} d x$ and $I_{2}=\frac{5}{6} \int \frac{1}{3 x^{2}+25 x+28} d x$

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

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

$\mathrm{As}, I_{1}=\frac{1}{6} \int \frac{(6 \mathrm{x}+25)}{3 \mathrm{x}^{2}+25 \mathrm{x}+28} \mathrm{dx}$

Let $u=3 x^{2}+25 x+28 \Rightarrow d u=(6 x+25) d x$

$\therefore I_{1}$ 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:

$I_{1}=\frac{1}{6} \log \left|3 x^{2}+25 x+28\right|+C \ldots . .$ eqn 2

As, $I_{2}=\frac{5}{6} \int \frac{1}{3 x^{2}+25 x+28} d x$ 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.

$\therefore I_{2}=\frac{5}{6} \int \frac{1}{3 x^{2}+25 x+28} d x=\frac{5}{6} \int \frac{1}{3\left(x^{2}+\frac{25}{3} x+\frac{28}{3}\right)} d x=\frac{5}{18} \int \frac{1}{x^{2}+\frac{25}{3} x+\frac{28}{3}} d x$

$\Rightarrow I_{2}=\frac{5}{18} \int \frac{1}{\left\{x^{2}+2\left(\frac{25}{6}\right) x+\left(\frac{25}{6}\right)^{2}\right\}+\frac{28}{3}-\left(\frac{25}{6}\right)^{2}} d x$

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

We have:

$I_{2}=\frac{5}{18} \int \frac{1}{\left(x+\frac{25}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}} d x$

$I_{2}$ matches with the form $\int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C$

$\therefore I_{2}=\frac{5}{18} \times \frac{1}{2 \times \frac{17}{6}} \log \left|\frac{\left(x+\frac{25}{6}\right)-\frac{17}{6}}{\left(x+\frac{25}{6}\right)+\frac{17}{6}}\right|+C$

$\therefore \mathrm{I}_{2}=\frac{5}{102} \log \left|\frac{6 \mathrm{x}+25-17}{6 \mathrm{x}+25+17}\right|+\mathrm{C}=\frac{5}{102} \log \left|\frac{6 \mathrm{x}-8}{6 \mathrm{x}+42}\right|+\mathrm{C} \ldots$ eqn 3

From eqn 1, we have:

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

Using eqn 2 and 3 , we get -

$I=\frac{1}{6} \log \left|3 x^{2}+25 x+28\right|+\frac{5}{102} \log \left|\frac{6 x-8}{6 x+42}\right|+C$