Evaluate, on intervals where the real-valued integrands are defined, \[ \int\frac{x^2+2x+1}{x\sqrt{x^2-1}}\,dx \] and \[ \int\frac{x+16}{\sqrt{x^2-4x+8}}\,dx. \]
Solution
For the first integral, split \[ \frac{x^2+2x+1}{x\sqrt{x^2-1}} =\frac{x}{\sqrt{x^2-1}}+\frac2{\sqrt{x^2-1}}+\frac1{x\sqrt{x^2-1}}. \] The three primitives are respectively \[ \sqrt{x^2-1},\qquad 2\ln|x+\sqrt{x^2-1}|, \qquad \arctan\sqrt{x^2-1}. \] Thus \[ \boxed{\sqrt{x^2-1}+2\ln|x+\sqrt{x^2-1}|+\arctan\sqrt{x^2-1}+C.} \]
For the second integral, write \[ x^2-4x+8=(x-2)^2+4, \qquad x+16=(x-2)+18. \] Then \[ \boxed{\int\frac{x+16}{\sqrt{x^2-4x+8}}\,dx =\sqrt{x^2-4x+8}+18\ln\!\left|x-2+\sqrt{x^2-4x+8}\right|+C.} \]