Question

Find \(\int \frac{3dx}{9+x^2}\).

(a) \(tan^{-1}⁡\frac{x}{2}+C\)

(b) \(tan^{-1}⁡\frac{x}{3}+C\)

(c) \(tan^{-1}\frac{x}{5}+C\)

(d) \(tan^{-1}⁡\frac{x}{4}+C\)

The question was asked in an internship interview.

My question is from Integrals of Some Particular Functions topic in portion Integrals of Mathematics – Class 12

Answer 1

Correct option is (b) \(tan^{-1}⁡\frac{x}{3}+C\)

The best I can explain: \(\int \frac{3dx}{9+x^2}=3\int \frac{dx}{3^2+x^2}\)

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

∴\(3\int \frac{dx}{x^2+3^2}=3\left (\frac{1}{3} tan^{-1}⁡\frac{x}{3}\right )+3C_1\)

\(3\int \frac{dx}{x^2+3^2}=tan^{-1}⁡\frac{x}{3}+C\).