I am not able to solve the integral

Question

$I=\int { \frac { \sqrt { \sqrt { x } +\sqrt { x-1 } } }{ 1+\sqrt { x } } }$ $J=\int { \frac { \sqrt { \sqrt { x } -\sqrt { x-1 } } }{ 1+\sqrt { x } } } $ then I-J

I substituted $x=\sec ^{ 2 }{ \theta } $.

$I-J=\frac { \sqrt { \sec { \theta } +\tan { \theta } } -\sqrt { \sec { \theta } -\tan { \theta } } }{ 1+\sec { \theta } } \times 2\sec ^{ 2 }{ \theta } \tan { \theta d\theta } \\ I-J=\frac { \sqrt { \sec { \theta } +\tan { \theta } } -\frac { 1 }{ \sqrt { \sec { \theta } +\tan { \theta } } } }{ 1+\sec { \theta } } \times 2\sec ^{ 2 }{ \theta } \tan { \theta d\theta } \\ I-J=\frac { \frac { \sec { \theta } +\tan { \theta } -1 }{ \sqrt { \sec { \theta } +\tan { \theta } } } }{ 1+\sec { \theta } } \times 2\sec ^{ 2 }{ \theta } \tan { \theta d\theta } $
I am not able to solve further Please help

Answer 1

There is some simplification available here as \begin{align} \left(\sqrt{\sqrt{x}+\sqrt{x-1}}\right)\left(\sqrt{\sqrt{x}-\sqrt{x-1}}\right) &= \sqrt{1} \end{align}

Since both $I$ and $J$ have these terms in, multiplication of these terms should give you something to work with once you have simplified the integrands using high school methods.

Answer 2

Hint:

$$\begin{align}I-J &=\int \frac{\sqrt{\sqrt{x}+\sqrt{x-1}}-\sqrt{\sqrt{x}-\sqrt{x-1}}}{\sqrt{x}+1}dx \\ &= \int\frac{\left(\left(\sqrt{\sqrt{x}+\sqrt{x-1}}-\sqrt{\sqrt{x}-\sqrt{x-1}}\right)^{2}\right)^{\frac{1}{2}}}{\sqrt{x}+1}dx \\ &=\sqrt{2} \int \frac{\sqrt{\sqrt{x}-1}}{\sqrt{x}+1}dx \end{align}$$