Stuck in integration: $\int {\frac{dx}{( 1+\sqrt {x})\sqrt{(x-{x}^2)}}}$

Question

$\displaystyle\int {\frac{dx}{( 1+\sqrt {x})\sqrt{(x-{x}^2)}}}$ $=\displaystyle\int\frac{(1-\sqrt x)}{(1+x)\sqrt{x-x^2}}\,dx$

$=\displaystyle\int\frac{(1-\sqrt x+x-x)}{(1+x)\sqrt{x-x^2}}\,dx$ $=\displaystyle\int\frac{\,dx}{\sqrt{x-x^2}}-\displaystyle\int\frac{\sqrt x(1+\sqrt x)}{\sqrt x(1+x)\sqrt{1-x}}\,dx$

I have tried the above integration a lot . But not able to solve further .

Please give me some hint how to proceed or some other method

Answer 1

By subtsitution twice we get $t=\sqrt { x } \Rightarrow dt=\frac { dx }{ 2\sqrt { x } } $ $$\int { \frac { dx }{ \left( 1+\sqrt { x } \right) \sqrt { x-{ x }^{ 2 } } } } =2\int { \frac { d\sqrt { x } }{ \left( 1+\sqrt { x } \right) \sqrt { 1-x } } = } 2\int { \frac { dt }{ \left( 1+t \right) \sqrt { 1-{ t }^{ 2 } } } } \\ \\ $$ $t=\sin { z } \Rightarrow dt=\cos { z } dz$

$$\int { \frac { \cos { z } dz }{ \left( 1+\sin { z } \right) \cos { z } } } =2\int { \frac { dz }{ 1+\sin { z } } } \overset { multiplying\quad both\quad (1-\sin { z) } \quad }{ = } 2\int { \frac { 1-\sin { z } }{ \cos ^{ 2 }{ z } } } dz=2\left( \int { \frac { dz }{ \cos ^{ 2 }{ z } } -\int { \frac { \sin { z } dz }{ \cos ^{ 2 }{ z } } } } \right) =2\left( \tan { z } -\frac { 1 }{ \cos { z } } \right) +C=\\ =2\left( \tan { \left( \arcsin { \left( \sqrt { x } \right) } \right) -\frac { 1 }{ \cos { \left( \arcsin { \left( \sqrt { x } \right) } \right) } } } \right) +C=2\left( \frac { \sqrt { x } }{ \sqrt { 1-x } } -\frac { 1 }{ \sqrt { 1-x } } \right) +C=2\frac { \sqrt { x } -1 }{ \sqrt { 1-x } } +C$$

Answer 2

Substitute $$1-x=t^2\implies dx=-2t\ dt$$

$$\int {dx\over \sqrt x(1+\sqrt x)\sqrt{1-x}}=\int {-2t\ dt\over\sqrt{1-t^2}(1+\sqrt{1-t^2})t} \\=-2\int\left({dt\over\sqrt{1-t^2}}-{dt\over 1+\sqrt{1-t^2}}\right)$$

Now can you solve the rest?

Answer 3

Putting x as (sin ø)^2 works as well.