Find the integral.
-$\int \frac{x}{(3 x+1)^{1 / 3}} d x$
A) $\frac{1}{3}\left[\frac{(3 x+1)^{5 / 3}}{5}-\frac{(3 x+1)^{2 / 3}}{2}\right]+C$
B) $\frac{(3 x+1)^{5 / 3}}{5}-\frac{(3 x+1)^{2 / 3}}{2}+C$
C) $\frac{1}{3} \frac{(3 x+1)^{5 / 3}}{5}-\frac{(3 x+1)^{2 / 3}}{2}+C$
D) $\frac{1}{3}\left[\frac{(3 x+1)^{5 / 3}}{2}-\frac{(3 x+1)^{2 / 3}}{5}\right]+C$

Question

Quizplus · Accepted Answer

The integral can be solved using substitution. Let $ u = (3x+1)^{1/3} $, then $ du = \frac{1}{3}(3x+1)^{-2/3} \cdot 3 \, dx = (3x+1)^{-2/3} \, dx $. Rewriting the integral in terms of $ u $ and integrating gives the result in option A.

Find the Integral ∫x(3x+1)1/3dx\int \frac{x}{(3 x+1)^{1 / 3}} d x∫(3x+1)1/3x​dx A) 13[(3x+1)5/35−(3x+1)2/32]+C\frac{1}{3}\left[\frac{(3 x+1)^{5 / 3}}{5}-\frac{(3 x+1)^{2 / 3}}{2}\right]+C31​[5(3x+1)5/3​−2(3x+1)2/3​]+C

Find the Integral $\int \frac{x}{(3 x+1)^{1 / 3}} d x$ A) $\frac{1}{3}\left[\frac{(3 x+1)^{5 / 3}}{5}-\frac{(3 x+1)^{2 / 3}}{2}\right]+C$