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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2019 · Mathematics · Indefinite Integration

What is ∫ (d x)/(2 x^2-2 x+1) equal to?

What is ∫dx2x2−2x+1\int \frac{d x}{2 x^2-2 x+1} equal to?

Options

  1. Atan⁡−1(2x−1)2+c\frac{\tan ^{-1}(2 x-1)}{2}+c
  2. B2tan⁡−1(2x−1)+c2 \tan ^{-1}(2 x-1)+c
  3. Ctan⁡−1(2x+1)2+c\frac{\tan ^{-1}(2 x+1)}{2}+c
  4. Dtan⁡−1(2x−1)+c\tan ^{-1}(2 x-1)+c

Correct answer

D. tan⁡−1(2x−1)+c\tan ^{-1}(2 x-1)+c

Step-by-step solution

∫dx2x2−2x+1=12∫dxx2−x+12=12∫dx(x−12)2+(12)2=12[112⋅tan⁡−1(x−1212)]+C=12⋅2:tan⁡−1(2x−1)+C=tan⁡−1(2x−1)+C\begin{aligned} & \int \frac{d x}{2 x^2-2 x+1}=\frac{1}{2} \int \frac{d x}{x^2-x+\frac{1}{2}} \\ & =\frac{1}{2} \int \frac{d x}{\left(x-\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2} \\ & =\frac{1}{2}\left[\frac{1}{\frac{1}{2}} \cdot \tan ^{-1}\left(\frac{x-\frac{1}{2}}{\frac{1}{2}}\right)\right]+C \\ & =\frac{1}{2} \cdot 2: \tan ^{-1}(2 x-1)+C \\ & =\tan ^{-1}(2 x-1)+C \end{aligned}
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