We will use u-substitute:[tex]u= \sqrt{2x} , \frac{du}{dx}= \frac{1}{ \sqrt{2x} }= \frac{1}{u} [/tex]Then for substitution:dx=u du. and integral becomes:[tex] \int { \frac{u}{u+1} } \, du = \int { \frac{u+1-1}{u+1} } \, du= \int{1} \, du- \int { \frac{1}{u-1} } \, dx [/tex]=u-ln(u+1)=[tex] \sqrt{2x}-ln( \sqrt{2x}+1) [/tex]. Now we will change the values of limits: [tex] \sqrt{16}-ln( \sqrt{16}+1)-( \sqrt{4}-ln( \sqrt{4}+1)) [/tex]=4-ln(5)-2+ln(3)=2+ln(0,6)=2-0.51=1.49
