Evaluate the following antiderivatives on intervals where the real-valued integrands are defined:
- \(\displaystyle \int \frac{1-\sqrt x}{1+\sqrt x}\,dx\).
- \(\displaystyle \int \frac{dx}{\sqrt x+2\sqrt[3]x}\).
- \(\displaystyle \int \frac{dx}{\sqrt x-\sqrt[3]x}\).
- \(\displaystyle \int \frac{dx}{\sqrt x+\sqrt[4]x}\).
- \(\displaystyle \int \frac{\sqrt[3]x+1}{\sqrt[3]x-1}\,dx\).
- \(\displaystyle \int \frac{dx}{(x+1)\sqrt x}\) and \(\displaystyle \int \frac{dx}{2x\sqrt{x-1}}\).
- \(\displaystyle \int \frac{x}{\sqrt[3]{x+1}}\,dx\).
- \(\displaystyle \int \frac{dx}{x\sqrt{4x-1}}\).
- \(\displaystyle \int \frac{2\,dx}{x\sqrt{25x^4-1}}\).
- \(\displaystyle \int \frac{dx}{\sqrt x\sqrt{1-4x}}\).
- \(\displaystyle \int \frac{6-2x}{\sqrt{9-x^2}}\,dx\).
- \(\displaystyle \int \frac{dx}{\sqrt{x\sqrt x-x^2}}\).
Solution
For (1), let \(t=\sqrt x\). Then \(dx=2t\,dt\) and \[ 2t\frac{1-t}{1+t}=-2t+4-\frac4{t+1}. \] Hence \[ \boxed{-x+4\sqrt x-4\ln(1+\sqrt x)+C.} \]
For (2), put \(t=x^{1/6}\). Then \[ \frac{dx}{\sqrt x+2\sqrt[3]x}=\frac{6t^3}{t+2}\,dt, \] so polynomial division gives \[ \boxed{2\sqrt x-6\sqrt[3]x+24\sqrt[6]x-48\ln|\sqrt[6]x+2|+C.} \]
For (3), with the same \(t=x^{1/6}\), \[ \frac{dx}{\sqrt x-\sqrt[3]x}=\frac{6t^3}{t-1}\,dt, \] thus \[ \boxed{2\sqrt x+3\sqrt[3]x+6\sqrt[6]x+6\ln|\sqrt[6]x-1|+C.} \]
For (4), let \(t=x^{1/4}\). Then \[ \frac{dx}{\sqrt x+\sqrt[4]x}=\frac{4t^2}{t+1}\,dt, \] so \[ \boxed{2\sqrt x-4\sqrt[4]x+4\ln|\sqrt[4]x+1|+C.} \]
For (5), let \(t=\sqrt[3]x\). Since \(dx=3t^2dt\), \[ 3t^2\frac{t+1}{t-1}=3t^2+6t+6+\frac6{t-1}, \] whence \[ \boxed{x+3x^{2/3}+6x^{1/3}+6\ln|x^{1/3}-1|+C.} \]
For the first integral in (6), let \(t=\sqrt x\): \[ \boxed{\int\frac{dx}{(x+1)\sqrt x}=2\arctan\sqrt x+C.} \] For the second, let \(t=\sqrt{x-1}\): \[ \boxed{\int\frac{dx}{2x\sqrt{x-1}}=\arctan\sqrt{x-1}+C.} \]
For (7), let \(u=x+1\): \[ \boxed{\frac35(x+1)^{5/3}-\frac32(x+1)^{2/3}+C.} \]
For (8), let \(u=\sqrt{4x-1}\). Then \[ \boxed{2\arctan\sqrt{4x-1}+C.} \]
For (9), let \(u=\sqrt{25x^4-1}\). Since \(u^2+1=25x^4\), the transformed integrand is \(du/(1+u^2)\), giving \[ \boxed{\arctan\sqrt{25x^4-1}+C.} \]
For (10), let \(u=2\sqrt x\). Then \(du=dx/\sqrt x\), so \[ \boxed{\arcsin(2\sqrt x)+C.} \] On \(0<x<1/4\) this is equivalent, up to an additive constant, to \(\tfrac12\arcsin(8x-1)\).
For (11), split the integral: \[ \int\frac6{\sqrt{9-x^2}}\,dx-\int\frac{2x}{\sqrt{9-x^2}}\,dx, \] which gives \[ \boxed{6\arcsin\!\left(\frac x3\right)+2\sqrt{9-x^2}+C.} \]
For (12), let \(u=x^{1/4}\). Since \[ \sqrt{x\sqrt x-x^2}=x^{3/4}\sqrt{1-\sqrt x}, \] and \(dx=4u^3du\), we obtain \[ \boxed{4\arcsin(x^{1/4})+C.} \]