<s>[INST] Integral $\sqrt{1+\frac{1}{4x}}$ $$\mathbf\int\sqrt{1+\frac{1}{4x}} \, dx$$
This integral came up while doing an arc length problem and out of curiosity I typed it into my TI 89 and got this output
$$\frac{-\ln(|x|)+2\ln(\sqrt\frac{4x+1}{x}-2)-2x\sqrt\frac{4x+1}{x}}{8}\ $$
How would one get this answer by integrating manually?
 [/INST] $$I=\int\frac{\sqrt{4x+1}}{2\sqrt{x}}dx$$   Use $$\sqrt{x}=t$$ $\implies$ $$\frac{dx}{2\sqrt{x}}=dt$$
we get  $$I=\int \sqrt{4t^2+1}dt=2\int\sqrt{t^2+0.5^2}dt$$
now use standard integral $$\int\sqrt{x^2+a^2}=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}ln|x+\sqrt{x^2+a^2}|$$
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