Choosing an upper limit to reach a given integral

Use the substitution \(u=1+x^2\) to show that \(\displaystyle\int_0^{\sqrt3}\dfrac{x}{\sqrt{1+x^2}}\,\mathrm dx=1\).[3]

Find the exact value of the positive constant \(k\) for which \(\displaystyle\int_0^{k}\dfrac{x}{\sqrt{1+x^2}}\,\mathrm dx=2\).[2]

Show that \(\dfrac{x}{\sqrt{1+x^2}}<1\) for all \(x\ge0\). Hence explain why \(\displaystyle\int_0^{k}\dfrac{x}{\sqrt{1+x^2}}\,\mathrm dx<k\) for every \(k>0\), even though the integral can be made as large as we like.[3]