how i should solve this problem?

Question

$\int \frac{\text{d}x}{(1+\sqrt{x})(x-x^2)}$

how i should solve this problem ? i think we should take $x=\sin^2(x)$ and then proceed but still not able to solve , please help !

Answer 1

HINT: Let $\sqrt x=y\implies x=y^2,dx=2y\ dy$

$$\int\dfrac{dx}{(1+\sqrt x)(x-x^2)}=\int\dfrac{2y}{(1+y)y^2(1-y^2)}dy$$

Partial Fraction Decomposition: $$\dfrac1{(1+y)^2y(1-y)}=\dfrac A{1-y}+\dfrac By+\dfrac C{1+y}+\dfrac D{(1+y)^2}$$

Answer 2

Using the substitution $x=\sin^2t$ we have the following: \begin{align*} \int\frac{dx}{(1+\sqrt{x})(x-x^2)}&=\int\frac{2\sin t\cos t \,dt}{(1+\sin t)(\sin^2 t)(1-\sin^2 t)}\\ &=\int\frac{2\,dt}{(1+\sin t)\sin t\cos t}\\ &=\int\frac{2(1-\sin t)\,dt}{\sin t\cos^3 t}\\ &=2\int (\tan t)^{-1}\sec^4 t\,dt-2\int \sec^3 t\,dt\\ &=2\int (\tan t)^{-1}(\tan^2 t + 1 )\sec^2 t\,dt-\sec t\tan t-\ln|\sec t+\tan t|\\ &=\tan^2 t+2\ln|\tan t|-\sec t\tan t-\ln|\sec t+\tan t|+C\\ &=\frac{x}{1-x}+\ln \left(\frac{x}{1-x}\right)-\frac{\sqrt{x}}{1-x}-\ln\left(\frac{1+\sqrt{x}}{\sqrt{1-x}}\right)+C\\ &=\ln\left[\frac{\sqrt{x}}{(1+\sqrt{x})\sqrt{1-x}}\right]-\frac{\sqrt{x}}{1+\sqrt{x}}+C \end{align*}