Question

$\int _ { 1 } ^ { 2 } \frac { x + 1 } { \sqrt { x - 1 } } d x$ is

• A. Convergent and equal to $\frac { 14 } { 3 }$
• B. Divergent and equal to $\frac { 3 } { 14 }$
• C. Convergent and equal to ∞
• D. Divergent and equal to ∞

Answer 1

Answer: (A)

Convergent and equal to $\frac { 14 } { 3 }$

Answer 2

$I = \int _ { 1 } ^ { 2 } \sqrt { x - 1 } d x + \int _ { 1 } ^ { 2 } \frac { 2 } { \sqrt { x - 1 } } d x$ = $\left[ \frac { 2 } { 3 } ( x - 1 ) ^ { 3 / 2 } \right] _ { 1 } ^ { 2 } + [ 4 \sqrt { x - 1 } ] _ { 1 } ^ { 2 }$

= $14 / 3$ which is finite so convergent.