The value of $∫\frac{dx}{x^2-6x+13}$

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

The value of $∫\frac{dx}{x^2-6x+13}$ is :

Options:

$\frac{1}{2}tan^{-1}\frac{x-3}{2}+C, $ where C is constant of integration.

$\frac{1}{2}cot^{-1}\frac{x-3}{2}+C, $ where C is constant of integration.

$\frac{1}{2}tan^{-1}\frac{x+3}{2}+C, $ where C is constant of integration.

$\frac{1}{2}cot^{-1}\frac{x+3}{2}+C, $ where C is constant of integration.

Correct Answer:

$\frac{1}{2}tan^{-1}\frac{x-3}{2}+C, $ where C is constant of integration.

Explanation:

The correct answer is Option (1) → $\frac{1}{2}\tan^{-1}\frac{x-3}{2}+C, $ where C is constant of integration.

$∫\frac{dx}{x^2-6x+13}$

$=∫\frac{dx}{(x-3)^2+2^2}$

$=\frac{1}{2}\tan^{-1}\frac{(x+3)}{2}+C$