Find the indefinite integral by making the substitution .

Question

Find the indefinite integral by making the substitution  . ​  ​

Quizplus · Accepted Answer

The given substitution involves a radical term, so it is natural to try to simplify the expression under the radical by completing the square. We have:

$$x^2-8x+20 = (x-4)^2-4$$

Making the substitution $u = x-4$ and using the identity $\cosh^2 t - \sinh^2 t = 1$, we get:

$$\begin{aligned} \int \sqrt{x^2-8x+20}\ dx &= \int \sqrt{(x-4)^2-4}\ dx \ &= \int \sqrt{4\cosh^2 	heta - 4}\ d	heta\ &= \int 2\sqrt{\cosh^2 	heta - 1}\cdot \cosh 	heta\ d	heta\ &= \int 2\sinh^2 	heta\ d	heta\ &= \int \sinh^2 	heta + \cosh 0\ d	heta \ &= \frac{1}{2}\sinh 	heta \cosh 	heta + 	heta + C \ &= \frac{1}{2}\sinh 	ext{arcosh}\left(\frac{x-4}{2}ight)\cosh 	ext{arcosh}\left(\frac{x-4}{2}ight) + 	ext{arcosh}\left(\frac{x-4}{2}ight) + C \ &= \frac{1}{2}\sqrt{(x-4)^2-4}\cdot \frac{x-4}{2} + 	ext{arcosh}\left(\frac{x-4}{2}ight) + C \ &= \frac{1}{4}(x-4)\sqrt{x^2-8x+20} + 	ext{arcosh}\left(\frac{x-4}{2}ight) + C \end{aligned}$$

Therefore, the correct answer is D.

Find the Indefinite Integral by Making the Substitution