Evaluate the integral:

$I=\int \frac{x+5}{3 x^{2}+13 x-10} d x$

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, $\frac{\mathrm{d}}{\mathrm{dx}}\left(3 \mathrm{x}^{2}+13 \mathrm{x}-10\right)=6 \mathrm{x}+13$

$\therefore$ Let, $x+5=A(6 x+13)+B$

$\Rightarrow x+5=6 A x+13 A+B$

On comparing both sides -

We have,

$6 A=1 \Rightarrow A=1 / 6$

$13 A+B=5 \Rightarrow B=-13 A+5=17 / 6$

Hence,

$I=\int \frac{\frac{1}{6}(6 x+13)+\frac{17}{6}}{3 x^{2}+13 x-10} d x$

$\therefore I=\int \frac{\frac{1}{6}(6 x+13)}{3 x^{2}+13 x-10} d x+\int \frac{\frac{17}{6}}{3 x^{2}+13 x-10} d x$

Let, $I_{1}=\frac{1}{6} \int \frac{(6 x+13)}{3 x^{2}+13 x-10} d x$ and $I_{2}=\frac{17}{6} \int \frac{1}{3 x^{2}+13 x-10} d x$

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

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

$\mathrm{As}, \mathrm{I}_{1}=\frac{1}{6} \int \frac{(6 \mathrm{x}+13)}{3 \mathrm{x}^{2}+13 \mathrm{x}-10} \mathrm{dx}$

Let $u=3 x^{2}+13 x-10 \Rightarrow d u=(6 x+13) d x$

$\therefore I_{1}$ reduces to $\frac{1}{6} \int \frac{\mathrm{du}}{\mathrm{u}}$

Hence,

$I_{1}=\frac{1}{6} \int \frac{d u}{u}=\frac{1}{6} \log |u|+C\left\{\because \int \frac{d x}{x}=\log |x|+C\right\}$

On substituting value of $u$, we have:

$I_{1}=\frac{1}{6} \log \left|3 x^{2}+13 x-10\right|+C \ldots .$ eqn 2

As, $I_{2}=\frac{17}{6} \int \frac{1}{3 x^{2}+13 x-10} d x$ 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.

$\therefore I_{2}=\frac{17}{6} \int \frac{1}{3 x^{2}+13 x-10} d x=\frac{17}{6} \int \frac{1}{3\left(x^{2}+\frac{12}{3} x-\frac{10}{3}\right)} d x=\frac{17}{18} \int \frac{1}{x^{2}+\frac{13}{3} x-\frac{10}{3}} d x$

$\Rightarrow I_{2}=\frac{17}{18} \int \frac{6}{\left\{x^{2}+2\left(\frac{12}{6}\right) x+\left(\frac{12}{6}\right)^{2}\right\}-\frac{10}{3}-\left(\frac{12}{6}\right)^{2}} d x$

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

We have:

$\mathrm{I}_{2}=\frac{17}{18} \int \frac{1}{\left(\mathrm{x}+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}} \mathrm{dx}$

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

$\therefore \mathrm{I}_{2}=\frac{1}{6} \log \left|\frac{6 \mathrm{x}+13-17}{6 \mathrm{x}+13+17}\right|+\mathrm{C}=\frac{1}{6} \log \left|\frac{6 \mathrm{x}-4}{6 \mathrm{x}+30}\right|+\mathrm{C} \ldots$ eqn 3

From eqn 1, we have:

$I=I_{1}+I_{2}$

Using eqn 2 and 3 , we get -

$I=\frac{1}{6} \log \left|3 x^{2}+13 x-10\right|+\frac{1}{6} \log \left|\frac{6 x-4}{6 x+30}\right|+C$