The definite integral $\int _ { 1 } ^ { 4 } \frac { 1 } { \sqrt { x } ( 1 + \sqrt { x } ) ^ { 2 } } d x$ is

A) $- \frac { 2 } { 3 }$
B) $- \frac { 1 } { 3 }$
C) $\frac { 4 } { 3 }$
D) $\frac { 1 } { 3 }$
E) $\frac { 3 } { 2 }$

Question

The definite integral $\int _ { 1 } ^ { 4 } \frac { 1 } { \sqrt { x } ( 1 + \sqrt { x } ) ^ { 2 } } d x$ is

A) $- \frac { 2 } { 3 }$
B) $- \frac { 1 } { 3 }$
C) $\frac { 4 } { 3 }$
D) $\frac { 1 } { 3 }$
E) $\frac { 3 } { 2 }$

Quizplus · Accepted Answer

To solve the integral $\int _ { 1 } ^ { 4 } \frac { 1 } { \sqrt { x } ( 1 + \sqrt { x } ) ^ { 2 } } d x$, let $u = \sqrt{x}$, which implies $x = u^2$ and $dx = 2u du$. Substituting these into the integral gives $\int _ { 1 } ^ { 2 } \frac { 2 } { ( 1 + u ) ^ { 2 } } du$, since the limits change to $u = 1$ to $u = 2$ when $x = 1$ and $x = 4$, respectively. This integral can be solved by recognizing it as a standard form, leading to the result of $\frac{1}{3}$.

The Definite Integral ∫141x(1+x)2dx\int _ { 1 } ^ { 4 } \frac { 1 } { \sqrt { x } ( 1 + \sqrt { x } ) ^ { 2 } } d x∫14​x​(1+x​)21​dx

The Definite Integral $\int _ { 1 } ^ { 4 } \frac { 1 } { \sqrt { x } ( 1 + \sqrt { x } ) ^ { 2 } } d x$