Question

Using the U- Substitution u=sqrt(2x), integral form 2-8 dx/ sqrt(2x) + 1 is equivalent to ...

Answer 1

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