Question

The value of integral $\int \frac{2x+4}{\sqrt{x^2-4x+5}}dx$ is equal to
(Where $C$ is integration constant)

• A. $\sqrt{x^2-4x+5}-8\ln |(x-2)+\sqrt{x^2-4x+5}|+C$
• B. $2\sqrt{x^2-4x+5}+8\ln |(x-2)+\sqrt{x^2-4x+5}|+C$
• C. $2\sqrt{x^2-4x+5}-8\ln |(x-2)+\sqrt{x^2-4x+5}|+C$
• D. $\frac{1}{2}\sqrt{x^2-4x+5}+8\ln |(x-2)+\sqrt{x^2-4x+5}|+C$

Answer 1

The correct option is B $2\sqrt{x^2-4x+5}+8\ln |(x-2)+\sqrt{x^2-4x+5}|+C$

Let $I=\int \frac{2x+4}{\sqrt{x^2-4x+5}}dx$
Now,
$2x+4=\lambda \frac{d}{dx}(x^2-4x+5)+\mu =\lambda (2x-4)+\mu$
Equating the coefficients of same degree on both sides, we get
$\lambda =1,\mu =8\Rightarrow I=\int \frac{(2x-4)+8}{\sqrt{x^2-4x+5}}dx=\int \frac{(2x-4)}{\sqrt{x^2-4x+5}}dx+8\int \frac{dx}{\sqrt{(x-2{)}^2+1}}=2\sqrt{x^2-4x+5}+8\ln |(x-2)+\sqrt{x^2-4x+5}|+C$