Given integral
We need to evaluate
I = ∫ d x ( 1 + x ) x − x 2 . I=\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}. I = ∫ ( 1 + x ) x − x 2 d x .
Simplify the square root in the denominator
Note that
x − x 2 = x ( 1 − x ) . x-x^2=x(1-x). x − x 2 = x ( 1 − x ) . So
x − x 2 = x ( 1 − x ) = x 1 − x . \sqrt{x-x^2}=\sqrt{x(1-x)}=\sqrt{x}\,\sqrt{1-x}. x − x 2 = x ( 1 − x ) = x 1 − x . Hence,
I = ∫ d x ( 1 + x ) x 1 − x . I=\int \frac{dx}{(1+\sqrt{x})\sqrt{x}\sqrt{1-x}}. I = ∫ ( 1 + x ) x 1 − x d x .
Substitute t = x t=\sqrt{x} t = x
Let
t = x ⟹ x = t 2 , d x = 2 t d t . t=\sqrt{x}\implies x=t^2,\quad dx=2t\,dt. t = x ⟹ x = t 2 , d x = 2 t d t . Then
1 − x = 1 − t 2 . \sqrt{1-x}=\sqrt{1-t^2}. 1 − x = 1 − t 2 . Substituting,
I = ∫ 2 t d t ( 1 + t ) t 1 − t 2 = ∫ 2 d t ( 1 + t ) 1 − t 2 . I=\int \frac{2t\,dt}{(1+t)t\sqrt{1-t^2}}
=\int \frac{2\,dt}{(1+t)\sqrt{1-t^2}}. I = ∫ ( 1 + t ) t 1 − t 2 2 t d t = ∫ ( 1 + t ) 1 − t 2 2 d t . Now use
1 − t 2 = ( 1 − t ) ( 1 + t ) . 1-t^2=(1-t)(1+t). 1 − t 2 = ( 1 − t ) ( 1 + t ) . Thus
( 1 + t ) 1 − t 2 = ( 1 + t ) ( 1 − t ) ( 1 + t ) = ( 1 + t ) 3 / 2 1 − t . (1+t)\sqrt{1-t^2}=(1+t)\sqrt{(1-t)(1+t)}=(1+t)^{3/2}\sqrt{1-t}. ( 1 + t ) 1 − t 2 = ( 1 + t ) ( 1 − t ) ( 1 + t ) = ( 1 + t ) 3/2 1 − t . So
I = ∫ 2 d t ( 1 + t ) ( 1 − t ) ( 1 + t ) = ∫ 2 d t 1 − t ( 1 + t ) 3 / 2 . I=\int \frac{2\,dt}{(1+t)\sqrt{(1-t)(1+t)}}
=\int \frac{2\,dt}{\sqrt{1-t}(1+t)^{3/2}}. I = ∫ ( 1 + t ) ( 1 − t ) ( 1 + t ) 2 d t = ∫ 1 − t ( 1 + t ) 3/2 2 d t .
Recognize a derivative form
Consider
f ( t ) = 1 − t 1 + t . f(t)=\sqrt{\frac{1-t}{1+t}}. f ( t ) = 1 + t 1 − t . Differentiate:
f ( t ) = ( 1 − t 1 + t ) 1 / 2 . f(t)=\left(\frac{1-t}{1+t}\right)^{1/2}. f ( t ) = ( 1 + t 1 − t ) 1/2 . Let
g ( t ) = 1 − t 1 + t . g(t)=\frac{1-t}{1+t}. g ( t ) = 1 + t 1 − t . Then
g ′ ( t ) = − ( 1 + t ) − ( 1 − t ) ( 1 + t ) 2 = − 2 ( 1 + t ) 2 . g'(t)=\frac{-(1+t)-(1-t)}{(1+t)^2}=\frac{-2}{(1+t)^2}. g ′ ( t ) = ( 1 + t ) 2 − ( 1 + t ) − ( 1 − t ) = ( 1 + t ) 2 − 2 . Therefore,
f ′ ( t ) = 1 2 ( 1 − t 1 + t ) − 1 / 2 ⋅ − 2 ( 1 + t ) 2 . f'(t)=\frac{1}{2}\left(\frac{1-t}{1+t}\right)^{-1/2}\cdot \frac{-2}{(1+t)^2}. f ′ ( t ) = 2 1 ( 1 + t 1 − t ) − 1/2 ⋅ ( 1 + t ) 2 − 2 . This simplifies to
f ′ ( t ) = − 1 1 − t ( 1 + t ) 3 / 2 . f'(t)=-\frac{1}{\sqrt{1-t}(1+t)^{3/2}}. f ′ ( t ) = − 1 − t ( 1 + t ) 3/2 1 . Hence,
I = ∫ 2 d t 1 − t ( 1 + t ) 3 / 2 = − 2 f ( t ) + C . I=\int \frac{2\,dt}{\sqrt{1-t}(1+t)^{3/2}}
=-2f(t)+C. I = ∫ 1 − t ( 1 + t ) 3/2 2 d t = − 2 f ( t ) + C . So
I = − 2 1 − t 1 + t + C . I=-2\sqrt{\frac{1-t}{1+t}}+C. I = − 2 1 + t 1 − t + C . Substituting back t = x t=\sqrt{x} t = x
I = − 2 1 − x 1 + x + C . I=-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C. I = − 2 1 + x 1 − x + C .
Match with the options
This is exactly Option B :
− 2 1 − x 1 + x + C . - 2\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}+C. − 2 1 + x 1 − x + C .
Verification with stored answer
Stored correct answer: B
Derived answer: B
They agree.