The value of the integral ∫ dx (1 + x ) x - x^ 2 is equal to (where, C & is the constant of integration)

Correct answer: C. 2( x - 1 1 - x ) + C Let, I = ∫ dx x (1 + x ) 1 - x Put, x = sin^ 2 θ and dx = 2sin θ cos θ ⇒ I = ∫ (2sin θ cos θ)/(sin θ(1 + sin θ)cos θ) = 2∫ (dθ)/(1 + sin θ) = 2∫ 1 - sin θ cos^ 2 θ dθ = 2(tan θ - sec θ) + C = 2( (x)/(1 - x) - 1 1 - x ) + C = 2( x - 1 1 - x ) + C

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The value of the integral ∫ dx (1 + x ) x - x^ 2 is equal to...

The value of the integral

∫ \dfrac{dx}{\left(1 + \sqrt{x}\right)\sqrt{x - x^{2}}}

is equal to (where,

C

& is the constant of integration)

Options

A $\dfrac{\sqrt{x} + 1}{\sqrt{1 - x}} + C$
B $2\left(\dfrac{\sqrt{x} - 1}{\sqrt{1 + x}}\right) + C$
C $2\left(\dfrac{\sqrt{x} - 1}{\sqrt{1 - x}}\right) + C$
D $\dfrac{\sqrt{x} + 1}{\sqrt{1 + x}} + C$

Correct answer

C. $2\left(\dfrac{\sqrt{x} - 1}{\sqrt{1 - x}}\right) + C$

Step-by-step solution

Let,

I = ∫ \dfrac{dx}{\sqrt{x}\left(1 + \sqrt{x}\right)\sqrt{1 - x}}

Put,

x = sin^{2}\theta

and

dx = 2sin \theta cos \thetad\theta

⇒ I = ∫ \dfrac{2sin \theta cos \thetad\theta}{sin \theta\left(1 + sin \theta\right)cos \theta}

= 2∫ \dfrac{d\theta}{1 + sin \theta} = 2∫ \dfrac{1 - sin \theta}{cos^{2}\theta}d\theta

= 2\left(tan \theta - sec \theta\right) + C

= 2\left(\sqrt{\dfrac{x}{1 - x}} - \dfrac{1}{\sqrt{1 - x}}\right) + C = 2\left(\dfrac{\sqrt{x} - 1}{\sqrt{1 - x}}\right) + C

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