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Home Questions P7TBYATXAJ
Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:

The value of $\int\frac{dx}{x^4+5x^2+4}$ is:

Options:

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

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

$\frac{1}{3}\tan^{-1}x+\frac{1}{6}\tan^{-1}(\frac{x}{2})+C$

$\frac{1}{3}\tan^{-1}x-\frac{1}{6}\tan^{-1}(\frac{x}{2})+C$

Correct Answer:

$\frac{1}{3}\tan^{-1}x-\frac{1}{6}\tan^{-1}(\frac{x}{2})+C$

Explanation:

$I=\int\frac{dx}{x^4+5x^2+4}=\int\frac{dx}{(x^2+1)(x^2+4)}=\int\frac{1}{3}[\frac{1}{x^2+1}-\frac{1}{x^2+4}]dx⇒I=\frac{1}{3}\tan^{-1}x-\frac{1}{6}\tan^{-1}(\frac{x}{2})+C$

CUET Questions