Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:

Evaluate $\int \sqrt{5 - 2x + x^2} dx$.

Options:

$\frac{x-1}{2}\sqrt{x^2 - 2x + 5} - 2 \ln|x - 1 + \sqrt{x^2 - 2x + 5}| + C$

$(x-1)\sqrt{x^2 - 2x + 5} + 4 \ln|x - 1 + \sqrt{x^2 - 2x + 5}| + C$

$\frac{x-1}{2}\sqrt{x^2 - 2x + 5} + 2 \ln|x - 1 + \sqrt{x^2 - 2x + 5}| + C$

$\frac{x-1}{2}\sqrt{x^2 - 2x + 5} + 2 \arcsin\left(\frac{x-1}{2}\right) + C$

Correct Answer:

$\frac{x-1}{2}\sqrt{x^2 - 2x + 5} + 2 \ln|x - 1 + \sqrt{x^2 - 2x + 5}| + C$

Explanation:

The correct answer is Option (3) → $\frac{x-1}{2}\sqrt{x^2 - 2x + 5} + 2 \ln|x - 1 + \sqrt{x^2 - 2x + 5}| + C$

Let $I = \int \sqrt{5 - 2x + x^2} dx = \int \sqrt{x^2 - 2x + 1 + 4} dx$

$= \int \sqrt{(x - 1)^2 + (2)^2} dx$

$= \frac{x - 1}{2} \sqrt{2^2 + (x - 1)^2} + \frac{2^2}{2} \log |x - 1 + \sqrt{2^2 + (x - 1)^2}| + C$

$= \frac{x - 1}{2} \sqrt{5 - 2x + x^2} + 2 \log |x - 1 + \sqrt{5 - 2x + x^2}| + C$