The integral $\int \frac { 3 x + 5 } { \sqrt { 3 x + 6 } } d x$ is

A) $\frac { 2 ( x + 1 ) \sqrt { 3 x + 6 } } { 3 } + C$
B) $\frac { 2 x ( x + 1 ) \sqrt { 3 x + 6 } } { 3 } + C$
C) $- \frac { 2 x ( x - 1 ) \sqrt { 3 x + 6 } } { 3 } + C$
D) $- \frac { 2 x ( x + 1 ) \sqrt { 3 x + 6 } } { 3 } + C$
E) $\frac { 2 x ( x - 2 ) \sqrt { 3 x + 6 } } { 3 } + C$

Question

The integral $\int \frac { 3 x + 5 } { \sqrt { 3 x + 6 } } d x$ is

A) $\frac { 2 ( x + 1 ) \sqrt { 3 x + 6 } } { 3 } + C$
B) $\frac { 2 x ( x + 1 ) \sqrt { 3 x + 6 } } { 3 } + C$
C) $- \frac { 2 x ( x - 1 ) \sqrt { 3 x + 6 } } { 3 } + C$
D) $- \frac { 2 x ( x + 1 ) \sqrt { 3 x + 6 } } { 3 } + C$
E) $\frac { 2 x ( x - 2 ) \sqrt { 3 x + 6 } } { 3 } + C$

Quizplus · Accepted Answer

To solve the integral $\int \frac { 3 x + 5 } { \sqrt { 3 x + 6 } } d x$, let $u = 3x + 6$, then $du = 3dx$. The integral becomes $\int \frac {u - 1}{\sqrt{u}} du$, which simplifies to $\int u^{1/2} - u^{-1/2} du$. Integrating gives $\frac{2}{3}u^{3/2} - 2u^{1/2} + C$, substituting back for $u$ gives $\frac{2}{3}(3x + 6)^{3/2} - 2(3x + 6)^{1/2} + C$. Simplifying this expression leads to the form $\frac { 2 ( x + 1 ) \sqrt { 3 x + 6 } } { 3 } + C$, which matches choice A.

The Integral ∫3x+53x+6dx\int \frac { 3 x + 5 } { \sqrt { 3 x + 6 } } d x∫3x+6​3x+5​dx

The Integral $\int \frac { 3 x + 5 } { \sqrt { 3 x + 6 } } d x$