Đáp án:
$I = ln(|x+1+\sqrt{(x+1)^2+4}|)+C\\$
$I = 2ln|\sqrt{x-1} + \sqrt{x+3}|+C$
Giải thích các bước giải:
$I = \int\limits^{}_{} {\frac{1}{\sqrt{(x+1)^2 + 4}}} \, dx \\ $
$t = x+1+\sqrt{(x+1)^2+4} \\
dt = 1 + \frac{2(x+1)}{2\sqrt{(x+1)^2+4}} dx \\
dt = \frac{\sqrt{(x+1)^2+4}+x+1}{\sqrt{(x+1)^2+4}} dx \\
dt = \frac{t}{\sqrt{(x+1)^2+4}} dx \\$
$I = \int\limits^{}_{} {\frac{1}{t}} \, dt\\$
$I = ln(|x+1+\sqrt{(x+1)^2+4}|)+C\\$
$I = \int\limits^{}_{} \frac{1}{\sqrt{(x-1)(x+3)}} \, dx\\$
$t = \sqrt{x-1} + \sqrt{x+3} \\
dt = \frac{\sqrt{x-1}+\sqrt{x+3}}{2\sqrt{(x-1)(x+3)}}dx \\
dt = \frac{t}{2\sqrt{(x-1)(x+3)}}dx$
$I = \int\limits^{}_{} {\frac{2}{t}} \, dt \\$
$I = 2ln|t|+C\\
I = 2ln|\sqrt{x-1} + \sqrt{x+3}|+C
$
