Question

Find the integrals of the following:

${\displaystyle \frac{1}{6x-7-x^2}}$

Answer 1

${\displaystyle \int \frac{1}{6x-7-x^2} \text{d}x=\int \frac{\text{d}x}{-7-x^2+6x}}$

= ${\displaystyle \int \frac{\text{d}x}{-7-(x^2-6x)}}$

= ${\displaystyle \int \frac{\text{d}x}{-7-[{(x-3)}^2-3^2]}}$

= ${\displaystyle \int \frac{\text{d}x}{-7-{(x-3)}^2+9}}$

= ${\displaystyle \int \frac{\text{d}x}{2-{(x-3)}^2}}$

= ${\displaystyle \int \frac{\text{d}x}{{\left(\sqrt{2}\right)}^2-{(x-3)}^2}}$

Put x – 3 = t, dx = dt

= ${\displaystyle \int \frac{\text{dt}}{({\sqrt{2}}^2-{\text{t}}^2)}}$

= ${\displaystyle \frac{1}{2\times \left(\sqrt{2}\right)}\log \left|\frac{\sqrt{2}+\text{t}}{\sqrt{2}-\text{t}}\right|+\text{c}}$

= ${\displaystyle \frac{1}{2\times \sqrt{2}}\log \left|\frac{\sqrt{2}+(x-3)}{\sqrt{2}-(x-3)}\right|+\text{c}}$