# 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

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