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

Category: QuestionsFind \int \frac{dx}{x^2-8x+20}.
Editor">Editor Staff asked 11 months ago

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
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1 Answers
Editor">Editor Staff answered 11 months ago

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