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

1 answer:

8 0

Let's apply u-substitution: $u= \sqrt{2x}, \frac{du}{dx}= \frac{1}{ \sqrt{2x} }= \frac{1}{u}$ In the substitution process:dx=u du. then the integral transforms to: $\int { \frac{u}{u+1} } \, du = \int { \frac{u+1-1}{u+1} } \, du= \int{1} \, du- \int { \frac{1}{u-1} } \, dx$ =u-ln(u+1)= $\sqrt{2x}-ln( \sqrt{2x}+1)$ . Next, we will adjust the 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

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