A question on indefinite integration... $$\int\frac{dx}{\left(\sqrt{x+\sqrt{x}}\right)^{99}}$$
I don't have any idea how to solve this question...I have tried rationalising the denominator, making the power 100 in the denominator and then integrating by part's etc...but couldn't solve further....
Please help me out......
 A: HINT, I set $n=99$ for your question:
$$\int\frac{1}{\left(\sqrt{x+\sqrt{x}}\right)^{n}}\space\text{d}x=\int\frac{1}{\left(\left(x+\sqrt{x}\right)^{\frac{1}{2}}\right)^{n}}\space\text{d}x=\int\frac{1}{\left(x+\sqrt{x}\right)^{\frac{n}{2}}}\space\text{d}x=$$

Substitute $u=\sqrt{x}$ and $\text{d}u=\frac{1}{2\sqrt{x}}\space\text{d}x$:

$$2\int\frac{1}{\left(u^2+u\right)^{\frac{n}{2}}}\space\text{d}u=2\int\frac{1}{\left(\left(u+\frac{1}{2}\right)^2-\frac{1}{4}\right)^{\frac{n}{2}}}\space\text{d}u=$$

Substitute $s=u+\frac{1}{2}$ and $\text{d}s=\text{d}u$:

$$2\int\frac{s-\frac{1}{2}}{\left(s^2-\frac{1}{4}\right)^{\frac{n}{2}}}\space\text{d}s=$$

Substitute $s=\frac{\sec(p)}{2}$ and $\text{d}s=\frac{\tan(p)\sec(p)}{2}\space\text{d}p$.
Then $\left(s^2-\frac{1}{4}\right)^{\frac{n}{2}}=\left(\frac{\sec^2(p)}{4}-\frac{1}{4}\right)^{\frac{n}{2}}=\frac{\tan^n(p)}{2^n}$ and $p=\text{arcsec}(2s)$:

$$2^n\int\cot^{n-2}(p)\csc(p)\left(\frac{\sec(p)}{2}-\frac{1}{2}\right)\space\text{d}p=\frac{2^n}{2}\int\cot^{n-1}(p)\sec(p)\left(\sec(p)-1\right)\space\text{d}p=$$
$$\frac{2^n}{2}\left[\int\left(\cot^{n-1}(p)\sec^2(p)-\cot^{n-1}(p)\sec(p)\right)\space\text{d}p\right]=$$
$$\frac{2^n}{2}\left[\int\cot^{n-1}(p)\sec^2(p)\space\text{d}p-\int\cot^{n-1}(p)\sec(p)\space\text{d}p\right]=$$
$$\frac{2^n}{2}\left[-\frac{\cot^{n-2}(p)}{n-2}-\int\cot^{n-1}(p)\sec(p)\space\text{d}p\right]=$$
$$\frac{2^n}{2}\left[-\frac{\cot^{n-2}(p)}{n-2}-\int\cot^{n-2}(p)\csc(p)\space\text{d}p\right]$$
A: Hints:
$$v^2 = x, 2v\text{ dv} = \text dx\\
\int {\text dx\over \left(\sqrt{x+\sqrt x}\right)^{99}} = \int{2v\text{ d}v\over \left(\sqrt {v^2+v}\right)^{99}}$$
$$v+\frac 12 = \frac 12\cosh \theta,\text{ dv}=\frac 12\sinh \theta\text{ d}\theta\\
\int{2v\text{ d}v\over \left(\sqrt {v^2+v}\right)^{99}}=\int{2v\text{ d}v\over \left(\sqrt {\left(v+\frac 12\right)^2-\frac 14}\right)^{99}}=\int{\frac 12(\cosh\theta-1)\sinh\theta\over \left(\sqrt{\frac 14\cosh^2\theta-\frac 14}\right)^{99}}\text d\theta\\
\cosh^2\theta - 1 = \sinh^2\theta\\
\to\int{\frac 12(\cosh\theta-1)\sinh\theta\over \frac 1{2^{99}}\sinh^{99}\theta}\text d\theta=2^{98}\int{\cosh\theta\over\sinh^{98}\theta}\text d\theta-2^{98}\int\frac {\text d\theta}{\sinh^{98}\theta}$$
Can you take it from here?
Other notes:
The substitution $v^2=x$ came about after trying others ($v=\sqrt{x+\sqrt x}$ for example).  Once the new form was obtained $\left({v\over\sqrt{av^2+bv+c}}\right)$ a trig or hyperbolic trig substitution seemed to make sense.  The relative simplicity of the hyperbolic trig substitution worked out pretty well; another option would be to use $v+\frac 12 = \frac 12\sec\theta$ and then $\sec^2\theta-1=\tan^2\theta$, but I never like dealing with the extra terms when it comes to taking the derivative of $\sec\theta$...
