Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{x^{2}+x-1}{x^{2}+x-6} d x$

Solution:

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

Expressing the integral $\int \frac{\mathrm{P}(\mathrm{x})}{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}} \mathrm{dx}=\int \mathrm{Q}(\mathrm{x}) \mathrm{dx}+\int \frac{\mathrm{R}(\mathrm{x})}{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}} \mathrm{dx}$

Let $x^{2}+x-1=x^{2}+x-6+5$

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

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

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

Consider $\int \frac{1}{x^{2}+x-6} d x$

Factorizing the denominator,

$\Rightarrow \int \frac{1}{x^{2}+x-6} d x=\int \frac{1}{(x-2)(x+3)} d x$

By partial fraction decomposition,

$\Rightarrow \frac{1}{(x-2)(x+3)}=\frac{A}{x-2}+\frac{B}{x+3}$

$\Rightarrow 1=A(x+3)+B(x-2)$

$\Rightarrow 1=A x+3 A+B x-2 B$

$\Rightarrow 1=(A+B) x+(3 A-2 B)$

$\Rightarrow$ Then $A+B=0 \ldots$ (1)

And $3 A-2 B=1 \ldots$ (2)

Solving (1) and (2),

$2 \times(1) \rightarrow 2 A+2 B=0$

$1 \times(2) \rightarrow 3 A-2 B=1$

$5 A=1$

$\therefore A=1 / 5$

Substituting A value in (1),

$\Rightarrow \mathrm{A}+\mathrm{B}=0$

$\Rightarrow 1 / 5+\mathrm{B}=0$

$\therefore B=-1 / 5$

Thus, $\frac{1}{(x-2)(x+3)}=\frac{1}{5(x-2)}-\frac{1}{5(x+3)}$

$=\frac{1}{5} \int \frac{1}{x-2} d x-\frac{1}{5} \int \frac{1}{x+3} d x$

Let $x-2=u \rightarrow d x=d u$

And $x+3=v \rightarrow d x=d v$.

$\Rightarrow \frac{1}{5} \int \frac{1}{u} d u-\frac{1}{5} \int \frac{1}{v} d v$

We know that $\int \frac{1}{x} d x=\log |x|+c$

$\Rightarrow \frac{1}{5} \log |\mathrm{u}|-\frac{1}{5} \log |\mathrm{v}|$

$\Rightarrow \frac{1}{5} \log |\mathrm{x}-2|-\frac{1}{5} \log |\mathrm{x}+3|$

$\Rightarrow \frac{1}{5}(\log |\mathrm{x}-2|-\log |\mathrm{x}+3|)$

Then,

$\Rightarrow 5 \int\left(\frac{1}{x^{2}+x-6}\right) d x+\int 1 d x=5\left(\frac{1}{5}(\log |x-2|-\log |x+3|)\right)+\int 1 d x$

We know that $\int 1 \mathrm{dx}=\mathrm{x}+\mathrm{c}$

$\Rightarrow(\log |x-2|-\log |x+3|)+x+c$

$\therefore I=\int \frac{x^{2}+x-1}{x^{2}+x-6} d x=-\log |x+3|+x+\log |x-2|+c$