Question

Find: ${\displaystyle \int \frac{1}{5+4x-x^2}dx}$

Answer 1

Let I = ${\displaystyle \int \frac{dx}{5+4x-x^2}}$

= ${\displaystyle \int \frac{dx}{5+4-4+4x-x^2}}$

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

= ${\displaystyle \frac{1}{2\times 3}\log \left|\frac{3+(x-2)}{3-(x-2)}\right|+c ...[\text{using}\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\log \left|\frac{a+x}{a-x}\right|+C]}$

= ${\displaystyle \frac{1}{6}\log \left|\frac{x+1}{5-x}\right|+C}$