What is ∫{dx}{2x²-2x+1} equal to?

Correct Answer is : tan−1(2x−1)+ctan^{-1} (2x - 1) + ctan−1(2x−1)+c

Correct Answer is : tan−1(2x−1)+ctan^{-1} (2x - 1) + ctan−1(2x−1)+c

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UPSC NDA 2 2019 Mathematics Solved Paper

Question : 81 of 120

Marks: +1, -0

What is

∫\frac{dx}{2x^{2}-2x+1}

equal to?

Solution:

Concept:

The integral of

\frac{1}{1+u^2}

\tan^{-1} u + C

Explanation:

Rewrite the denominator using completing the square:

2x^2 - 2x + 1 = \frac{1}{2}\big[(4x^2 - 4x + 2)\big] = \frac{1}{2}\big[(4x^2 - 4x + 1) + 1\big] = \frac{1}{2}\big[(2x-1)^2 + 1\big]

Thus,

\frac{1}{2x^2 - 2x + 1} = \frac{2}{(2x-1)^2 + 1}

So the integral becomes

\int \frac{2\,dx}{(2x-1)^2 + 1}

Let

t = 2x - 1

. Then

dt = 2\,dx

Substitute:

\int \frac{2\,dx}{(2x-1)^2 + 1} = \int \frac{dt}{t^2 + 1}

Using the standard formula:

\int \frac{dt}{t^2+1} = \tan^{-1} t + C

Replace

t

with

2x-1

\tan^{-1}(2x-1) + C

Answer:

\tan^{-1}(2x-1) + C

, which corresponds to option D.

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