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

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Question: Answered & Verified by Expert

The value of the integral

\int \frac{x^{2} - 4 x \sqrt{x} + 6 x - 4 \sqrt{x} + 1}{x - 2 \sqrt{x} + 1} d x

is equal to

MathematicsIndefinite IntegrationJEE Main

Options:

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

Solution:

1077 Upvotes Verified Answer

The correct answer is:

\frac{x^{2}}{2} - \frac{4}{3} x^{\frac{3}{2}} + x + c

Ι = \int \frac{{(\sqrt{x} - 1)}^{4}}{{(\sqrt{x} - 1)}^{2}} d x = \int {(\sqrt{x} - 1)}^{2} d x = \int (x - 2 \sqrt{x} + 1) d x

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

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