Question

Find \(\int \frac{dx}{x^2-8x+20}\).

(a) \(\frac{1}{2} tan^{-1}⁡\frac{x^2-8x}{2}+C\)

(b) \(\frac{5}{2} tan^{-1}⁡\frac{x-4}{2}+C\)

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

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

I got this question in exam.

This intriguing question originated from Integrals of Some Particular Functions topic in chapter Integrals of Mathematics – Class 12

Answer 1

Right answer is (c) \(\frac{1}{2} tan^{-1}⁡\frac{x-4}{2}+C\)

Explanation: \(\int \frac{dx}{x^2-8x+20}=\int \frac{dx}{(x^2-2(4x)+4^2)+4}\)

=\(\int \frac{dx}{(x-4)^2+2^2}\)

Let x-4=t

Differentiating w.r.t x, we get

dx=dt

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

\(\int \frac{dx}{(x-4)^2+2^2}=\int \frac{dt}{t^2+2^2}=\frac{1}{2} tan^{-1}⁡\frac{t}{2}+C\)

Replacing t with x-4, we get

\(\int \frac{dx}{(x-4)^2+2^2}=\frac{1}{2} tan^{-1}⁡\frac{x-4}{2}+C\)