Question

Calculate the following integral ∫ (√x+ 1÷√x)^2 dx

Answer 1

Answer:

$\int\limits (\sqrt{x}+\frac{1}{\sqrt{x}})^{2} dx\\ =\int\limits (x+\frac{1}{x}+2) dx\\ =\int\limits x\: dx+\int\limits \frac{1}{x}\: dx+\int\limits 2\: dx\\ =\frac{x^{2}}{2}+C_{1}+log_{e}x+C_{2}+2x+C_{3}\\ =\frac{x^{2}}{2}+log_{e}x+2x+C\: \: \:(where C=C_{1}+C_{2}+C_{3},\: is\: another\: arbitrary\: constant\: of\: integration)\\ =\frac{x^{2}}{2}+2x+log_{e}x+C$