Integral Calculus of Rational Functions

Question

Using this rule: Integral Calculus "Rational Function Rule"

$$\int\frac{1}{a^{2}-x^{2}}\,dx = \frac{1}{2a}\log_{e}\left(\frac{a+x}{a-x}\right)+c$$

I have integrated a function, but my question is if I'm solving for $x$, in the next line I must raise both sides to the base of $e$, and there is still $+c$ (constant) on the end of the integrated function, so do I leave it as plus $c$ on the RHS on the equation (not do anything to it, so when I want to work it out by substituting $(x_1,y_1)$ at the end it will be $x_1 = \dotsc y_1\dotsc +c$) or when I raise both sides to the base of $e$ does $c$, the constant become $c*e\dotsc$ instead of $+c$?

Any help would be great.

Answer 1

If $y=x+C$, then $e^{y}= e^{x+C} = e^x e^{C}$, so the additive arbitrary constant becomes a multiplicative arbitrary constant.

Answer 2

The antiderivative exists in each of the follwing intervals :

$$(-\infty,-a);(-a,a);(a,+\infty) $$

the result should be

$$\frac12\ln (\frac {\frac{a+x}{a-x}}{\lambda} )$$ $$=\frac12(\ln ( |\frac {a+x}{a-x}| )-\ln (|\lambda|) $$

$$=\frac 12\ln (|\frac{a+x}{a-x}|)+C $$