Use the substitution $w=\sqrt{x}$ to transforms the integral $\int_{0}^{4} \frac{1}{x} \ln (1+\sqrt{x}) d x$
A. $\frac{1}{2} \int_{0}^{2} \frac{1}{w} \ln (1+w) d w$
B. $2 \int_{0}^{2} \frac{1}{w} \ln (1+w) d w$
C. $\frac{1}{2} \int_{0}^{2} \frac{1}{w^{2}} \ln (1+w) d w$
D. $\frac{1}{2} \int_{0}^{4} \frac{1}{w^{2}} \ln (1+w) d w$
E. $\frac{1}{2} \int_{0}^{4} w \ln (1+w) d w$
F. $\frac{1}{2} \int_{0}^{2} w \ln (1+w) d w$
Answer
\(\boxed{\text{The correct answer is A. } \frac{1}{2} \int_{0}^{2} \frac{1}{w} \ln (1+w) d w}\)
Step 1 ：Given the integral \(\int_{0}^{4} \frac{1}{x} \ln (1+\sqrt{x}) d x\)
Step 2 ：Use the substitution \(w=\sqrt{x}\) to transform the integral. This implies \(x=w^2\) and \(dx=2wdw\)
Step 3 ：When \(x=0\), \(w=0\) and when \(x=4\), \(w=2\). So, the limits of the integral change from 0 to 2
Step 4 ：Substituting these into the integral, we get \(\frac{1}{2} \int_{0}^{2} \frac{1}{w} \ln (1+w) d w\)
Step 5 ：\(\boxed{\text{The correct answer is A. } \frac{1}{2} \int_{0}^{2} \frac{1}{w} \ln (1+w) d w}\)
