Find the indefinite integral. $$\int \frac { w } { \sqrt { 1 + 3 w } } d w$$
A)$\frac { ( 3 w - 2 ) \sqrt { 3 w + 1 } } { 9 } + C$
B)$\frac { 2 ( 3 w - 2 ) \sqrt { 3 w + 1 } } { 9 } + C$
C) $\frac { 2 ( 3 w - 2 ) \sqrt { 3 w + 1 } } { 27 } + C$
D) $\frac { ( 3 w - 2 ) \sqrt { 3 w + 1 } } { 18 } + C$
E) $\frac { 2 ( 2 - 3 w ) \sqrt { 3 w + 1 } } { 27 } + C$

Question

Find the indefinite integral. $$\int \frac { w } { \sqrt { 1 + 3 w } } d w$$
A)$\frac { ( 3 w - 2 ) \sqrt { 3 w + 1 } } { 9 } + C$ 
B)$\frac { 2 ( 3 w - 2 ) \sqrt { 3 w + 1 } } { 9 } + C$
C) $\frac { 2 ( 3 w - 2 ) \sqrt { 3 w + 1 } } { 27 } + C$
D) $\frac { ( 3 w - 2 ) \sqrt { 3 w + 1 } } { 18 } + C$
E) $\frac { 2 ( 2 - 3 w ) \sqrt { 3 w + 1 } } { 27 } + C$

Quizplus · Accepted Answer

The integral can be solved using substitution. Let $ u = \sqrt{1 + 3w} $, then $ du = \frac{3}{2\sqrt{1+3w}} dw $. Solving the integral gives $\frac{2(3w-2)\sqrt{3w+1}}{27} + C$.

Find the Indefinite Integral ∫w1+3wdw\int \frac { w } { \sqrt { 1 + 3 w } } d w∫1+3w​w​dw

Find the Indefinite Integral $\int \frac { w } { \sqrt { 1 + 3 w } } d w$