Integration of $\int\log(\sqrt{1-x}+\sqrt{1+x}) \, dx$

Question

Integration of

$$\int\log\left(\sqrt{1-x}+\sqrt{1+x}\right) \, dx$$

Please help to go through this problem as i have started with putting $x$= $cos2y$.

Answer 1

It is shorter to integrate by parts, setting $ u = \ln\bigl(\sqrt{1-x} + \sqrt{1 + x}\bigr) $, hence: \begin{align*} \mathrm d\mkern1.5mu u & = \frac{\frac12\Bigl(\dfrac{-1}{\sqrt{1-x}} + \dfrac{1}{\sqrt{1 + x}}\Bigr)}{\sqrt{1-x} + \sqrt{1 + x}}\,\mathrm d\mkern1.5mu x= \frac12\frac{\sqrt{1-x}-\sqrt{1 + x}}{\sqrt{(1-x²)}\Bigl(\sqrt{1-x} + \sqrt{1 + x}\Bigr)}\,\mathrm d\mkern1.5mu x\\& = -\frac{1}{4x}\frac{\bigl(\sqrt{1-x}-\sqrt{1 + x}\bigr)² }{\sqrt{(1-x²)}}\,\mathrm d\mkern1.5mu x =\frac{1}{2x}\frac{\sqrt{1-x²}-1}{\sqrt{1-x²}}\,\mathrm d\mkern1.5mu x \end{align*} Setting $I=\displaystyle\int\log(\sqrt{1-x}+\sqrt{1+x})\,\mathrm d\mkern1.5mu x$, you get: \begin{align*} I & =x\ln\bigl(\sqrt{1-x}+\sqrt{1+x}\bigr)-\frac{1}{2}\int\Bigl(1-\frac{1}{\sqrt{1-x²}}\Bigr)\,\mathrm d\mkern1.5mu x\\ & = x \ln\bigl(\sqrt{1-x}+\sqrt{1+x}\bigr)+\frac12(\arcsin x-x). \end{align*}

Answer 2

hint: $x = \sin (2u)$. Can you complete square and take square root? and don't forget the integration by part.

Answer 3

Consider the integral \begin{align} I = \int\log(\sqrt{1-x}+\sqrt{1+x})dx \end{align} It is seen that \begin{align} I &= \int \ln\left[ \sqrt{1 -x} \, \left(1 + \sqrt{\frac{1+x}{1-x}} \right) \right] dx \\ &= \frac{1}{2} \int \ln(1-x) \, dx + \int \ln\left(1 + \sqrt{\frac{1+x}{1-x}} \right) dx \end{align} Now make the substitution \begin{align} u^{2} = \frac{1+x}{1-x} \end{align} to obtain \begin{align} I &= \frac{1}{2} [ (x-1) \ln(1-x) - x] + \int \frac{4 \ln(1+u) \, u \, du}{(1+u^{2})^{2}}. \end{align} Integration by parts leads to \begin{align} I &= \frac{1}{2} [ (x-1) \ln(1-x) - x] + \frac{2 \ln(1+u)}{1+u^{2}} - \int \frac{du}{(1+u)(1+u^{2})} \\ &= \frac{1}{2} [ (x-1) \ln(1-x) - x] + \frac{2 \ln(1+u)}{1+u^{2}} + \frac{1}{4} \ln(1 + u^{2}) - \frac{1}{2} \ln(1+u) - \frac{1}{2} \tan^{-1}(u). \end{align} Now reverse substitution yields \begin{align} I &= \frac{1}{2} [ (x-1) \ln(1-x) - x] + \left(\frac{1}{2}-x\right) \ln\left(1+ \sqrt{\frac{1 + x}{1-x}} \right) + \frac{1}{4} \ln\left(\frac{2}{1-x} \right) - \frac{1}{2} \tan^{-1}\left( \sqrt{\frac{1 + x}{1-x}} \right) \\ &= \frac{\ln(2)}{4} - \frac{x}{2} + \frac{1}{2} \left( x - \frac{3}{2}\right) \ln(1-x) + \left(\frac{1}{2}-x\right) \ln\left(1+ \sqrt{\frac{1 + x}{1-x}} \right) - \frac{1}{2} \tan^{-1}\left( \sqrt{\frac{1 + x}{1-x}} \right) \end{align}