The value of the integral ∫ x^ 2 - 4x x + 6x - 4 x + 1 x - 2 x + 1 dx & is equal to

Correct answer: B. x^ 2 2 - (4)/(3)x^ (3)/(2) + x + c Ι = ∫ ( x - 1) ^ 4 ( x - 1) ^ 2 dx = ∫ ( x - 1) ^ 2 dx = ∫ (x - 2 x + 1)dx = x^ 2 2 - (4)/(3)x^ (3)/(2) + 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 ∫ x^ 2 - 4x x + 6x - 4 x + 1 x - 2 x + 1...

The value of the integral

∫ \dfrac{x^{2} - 4x\sqrt{x} + 6x - 4\sqrt{x} + 1}{x - 2\sqrt{x} + 1}dx

& is equal to

Options

A $\dfrac{x^{\dfrac{3}{2}}}{2} + x + c$
B $\dfrac{x^{2}}{2} - \dfrac{4}{3}x^{\dfrac{3}{2}} + x + c$
C $x^{\dfrac{3}{2}} + \dfrac{x}{2} + c$
D $\dfrac{2}{3}x^{\dfrac{3}{2}} + c$

Correct answer

B. $\dfrac{x^{2}}{2} - \dfrac{4}{3}x^{\dfrac{3}{2}} + x + c$

Step-by-step solution

Ι = ∫ \dfrac{{\left(\sqrt{x} - 1\right)}^{4}}

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

<br/>

= \dfrac{x^{2}}{2} - \dfrac{4}{3}x^{\dfrac{3}{2}} + x + c

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