1 problem found
The variables \(t\) and \(x\) are related by \(t=x+ \sqrt{x^2+2bx+c\;} \,\), where \(b\) and \(c\) are constants and \(b^2 < c\). Show that \[ \frac{\d x}{\d t} = \frac{t-x}{t+b}\;, \] and hence integrate \(\displaystyle \frac1 {\sqrt{x^2+2bx+c}}\,\). Verify by direct integration that your result holds also in the case \(b^2=c\) if \(x+b > 0\) but that your result does not hold in the case \(b^2=c\) if \(x+b < 0\,\).
Solution: \begin{align*} && t &= x+ \sqrt{x^2+2bx+c} \\ && \frac{\d t}{\d x} &= 1 + \frac{x+b}{\sqrt{x^2+2bx+c}} \\ &&&= \frac{x + \sqrt{x^2+2bx+c} + b}{\sqrt{x^2+2bx+c}} \\ &&&= \frac{t+b}{t-x} \\ \Rightarrow && \frac{\d x}{\d t} &= \frac{t-x}{t+b} \\ \\ && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= \int \frac{1}{t-x} \frac{t-x}{t+b} \d t \\ &&&= \int \frac{1}{t+b} \d t \\ &&&= \ln (t + b) +C \\ &&&= \ln \left (x + \sqrt{x^2+2bx+c} + b \right) + C \end{align*} If \(b^2 = c\) then we have \(x^2+2bx+b^2 = (x+b)^2\) so \(\sqrt{x^2+2bx+c^2} = x+b\) (if \(x+b>0\)), so \begin{align*} && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= \int \frac{1}{x+b} \d x\\ &&&= \ln (x + b) + C \\ &&&= \ln(x+b) + \ln 2 + C' \\ &&&= \ln (2(x+b)) + C' \\ &&&= \ln \left(x + b + \sqrt{(x+b)^2} \right)+C'\\ &&&= \ln \left(x + b + \sqrt{x^2+2bx+c} \right)+C'\\ \end{align*} If \(x+b < 0\) then the antiderivative is \(\ln 0\). \begin{align*} && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= -\int \frac{1}{x+b} \d x\\ &&&= -\ln |x + b| + C \\ \end{align*} which are clearly different.