Question

Find \(\int \frac{dx}{x^2+4}\).

(a) –\(tan^{-1}\frac{⁡x}{4}+C\)

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

(c) \(\frac{3}{4} tan^{-1}⁡x+C\)

(d) \(\frac{3}{4} tan^{-1}\frac{⁡3x}{2}+C\)

This question was posed to me by my school teacher while I was bunking the class.

My doubt stems from Integrals of Some Particular Functions in chapter Integrals of Mathematics – Class 12

Answer 1

Correct answer is (b) \(\frac{1}{2} tan^{-1}⁡\frac{x}{2}+C\)

The explanation: \(\int \frac{dx}{x^2+4}=\int \frac{dx}{x^2+2^2}\)

Using the formula \(\int \frac{dx}{a^2+x^2}=\frac{1}{a} tan^{-1}⁡\frac{x}{a}+C\)

∴\(\int \frac{dx}{x^2+2^2}=(\frac{1}{2} tan^{-1}⁡\frac{x}{2})+C\)