Question

Integrate $\underset{0}{\overset{a}{\int }}\frac{dx}{ax+a^2-x^2}$

Answer 1

Solution :

$I={\int }_0^a\frac{dx}{ax+a^2-x^2}$

$={\int }_0^a\frac{dx}{a^2-(a{x}^2-ax)}$

$={\int }_0^a\frac{dx}{a^2-\left[\right(x-\frac{1}{2}a{)}^2]-\frac{x^2}{4}}$

$={\int }_0^a\frac{dx}{a^2-\frac{a^2}{4}-\left[\right(x-\frac{a}{2}{)}^2]}$

$={\int }_0^a\frac{dx}{\frac{5a^2}{4}-\left[\right(x-a/2{)}^2]}$

$={\int }_0^a\frac{dx}{(\frac{\sqrt{5}a}{2}{)}^2-[(x-\frac{a}{2}{)}^2]}$

$=\frac{1}{2\frac{a\sqrt{5}}{2}}ln|\frac{x-\frac{a}{2}-\frac{\sqrt{5}}{2}a}{x-\frac{a}{2}-\frac{\sqrt{5}a}{2}}|$

$=[\frac{\sqrt{5}}{a}ln|\frac{x-a/2+\frac{\sqrt{5}}{2}a}{x-\frac{a}{2}-\frac{\sqrt{5}}{2}a}|{]}_0^a$

$=\frac{\sqrt{5}}{a}ln|\frac{a/2+\sqrt{5}/2a}{a/2-\sqrt{5}/2a}|-\frac{\sqrt{5}}{a}ln|\frac{-a/2+\sqrt{5}/2a}{-a/2-\sqrt{5}/2a}|$

$=\frac{\sqrt{5}}{a}ln|\frac{a+\sqrt{5}a}{a-\sqrt{5}a}|-\frac{\sqrt{5}}{a}ln|\frac{-a+\sqrt{5}a}{-a-\sqrt{5}a}$

$=\frac{\sqrt{5}}{a}[ln|\frac{1+\sqrt{5}}{1-\sqrt{5}}|-\frac{\sqrt{-1+\sqrt{5}}}{-1-\sqrt{5}}|]$