1. For $\int \sqrt{x}\,\ln x\,dx$, which IBP assignment is correct and what does the remaining integral simplify to?
- A.$u = \ln x$, $dv = x^{1/2}dx$; remaining integral is $\int \frac{2}{3}x^{1/2}\,dx$
- B.$u = \sqrt{x}$, $dv = \ln x\,dx$; remaining integral is $\int \frac{x\ln x - x}{2\sqrt{x}}\,dx$
- C.$u = \ln x$, $dv = x^{1/2}dx$; remaining integral is $\int \frac{2}{3}x^{3/2}\,dx$
- D.$u = x^{1/2}\ln x$, $dv = dx$; remaining integral is $\int \frac{1}{\sqrt{x}}\,dx$
View Answer
Answer: $u = \ln x$, $dv = x^{1/2}dx$; remaining integral is $\int \frac{2}{3}x^{1/2}\,dx$
Set up IBP using LIATE: $\ln x$ is logarithmic (L), highest LIATE priority. Let $u = \ln x$, $dv = x^{1/2}\,dx$. Then $du = \frac{1}{x}\,dx$, $v = \frac{2}{3}x^{3/2}$. Compute the remaining integral: $\int v\,du = \int \frac{2}{3}x^{3/2} \cdot \frac{1}{x}\,dx = \int \frac{2}{3}x^{1/2}\,dx$. This matches Option A. Why distractors fail: Option B reverses the LIATE priority and produces a messy integral. Option C incorrectly simplifies $x^{3/2}/x$ as $x^{3/2}$ instead of $x^{1/2}$. Option D uses a non-standard grouping.