MCQ
The value of $\int_2^3 \frac{x+1}{x^2(x-1)} d x$ is
- A$2 \log 2-\frac{1}{6}$
- ✓$\log \frac{16}{9}-\frac{1}{6}$
- C$\log \frac{4}{3}-\frac{1}{6}$
- D$\log \frac{16}{9}+\frac{1}{6}$
✓
Answer
Correct option: B.
$\log \frac{16}{9}-\frac{1}{6}$
(B)
$\int_2^3 \frac{x+1}{x^2(x-1)} d x=\int_2^3\left(-\frac{1}{x^2}-\frac{2}{x}+\frac{2}{x-1}\right) d x$
$\begin{array}{l}=\left[\frac{1}{x}\right]_2^3-2[\log x]_2^3+2[\log (x-1)]_2^3 \\ =\frac{1}{3}-\frac{1}{2}-2 \log \frac{3}{2}+2 \log 2 \\ =\log \frac{16}{9}-\frac{1}{6}\end{array}$
$\int_2^3 \frac{x+1}{x^2(x-1)} d x=\int_2^3\left(-\frac{1}{x^2}-\frac{2}{x}+\frac{2}{x-1}\right) d x$
$\begin{array}{l}=\left[\frac{1}{x}\right]_2^3-2[\log x]_2^3+2[\log (x-1)]_2^3 \\ =\frac{1}{3}-\frac{1}{2}-2 \log \frac{3}{2}+2 \log 2 \\ =\log \frac{16}{9}-\frac{1}{6}\end{array}$
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