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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»

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

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


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