Integrate $\ln(\sqrt{1-x}+\sqrt{1+x})$

Question

Integrate $$\ln(\sqrt{1-x}+\sqrt{1+x})$$

---

My attempts:

i) Multiplying and dividing its conjugate, we get $$I=\int \ln(-2x)dx-\int\ln(\sqrt{1-x}-\sqrt{1+x})dx$$

The second term is equally challenging as the original one, so its of no use.

ii) $\sqrt{1-x}+\sqrt{1+x}=t \implies\frac{1}{2}(\frac{1}{\sqrt{1+x}}-\frac{1}{\sqrt{1-x}})dx=dt$

Though $\sqrt{1-x}-\sqrt{1+x}$ can be written in terms of $t$, nothing can be done to $\sqrt{1-x^2}$

iii) Let $x=\cos(2\theta) \implies dx=-2\sin(2\theta)d\theta$

$$I=\ln(\sqrt2(|\sin(\theta)|+|\cos(\theta)|))\cdot-2\sin(2\theta)d\theta$$

$$I=\ln(2(2\sin(\theta)\cos(\theta)+1))\cdot-\sin(2\theta)d\theta$$

$$I=\ln(2(\sin(2\theta)+1))\cdot-\sin(2\theta)d\theta$$

$y=2(1+\sin(2\theta)) \implies dy=4\cos(2\theta)d\theta$

$$I=-\ln(y)\cdot (\frac{y}{2}-1)\cdot \frac{1}{4\sqrt{y-\frac{y^2}{4}}} dy$$

Which doesnt seem promising.

iv) Using IBP(as suggested by @AnginaSeng), $$I=x\ln(\sqrt{1-x}+\sqrt{1+x})-\int x \frac{1}{\sqrt{1-x}+\sqrt{1+x}}\cdot \frac{1}{2}\cdot \frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{1-x^2}}$$

$$I=x\ln(\sqrt{1-x}+\sqrt{1+x})-\frac{1}{2}\int \frac{x}{\sqrt{1-x^2}} \frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{1-x}+\sqrt{1+x}}$$

---

A new method, or some hints for existing ones is appreciated

Answer 1

I think IBP approach you did (as @AnginaSeng said) is on the right track. We have $$ \frac{x}{\sqrt{1-x^{2}}}\frac{\sqrt{1-x} - \sqrt{1+x}}{\sqrt{1-x} +\sqrt{1+x}} = \frac{x}{\sqrt{1-x^{2}}} \frac{(\sqrt{1-x} - \sqrt{1+x})^{2}}{-2x} =- \frac{2 - 2\sqrt{1-x^{2}}}{2\sqrt{1-x^{2}}} = 1 - \frac{1}{\sqrt{1-x^{2}}} $$ and I bet you that you can deal with this integral.