Chapter 6, Problem 12RE

To determine

To find: the given integral.

Answer to Problem 12RE

  ${\displaystyle \int \frac{dx}{\sqrt[3]{3x+4}}}=\frac{1}{2}{\left(3x+4\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+C$ .

Explanation of Solution

Given:

  ${\displaystyle \int \frac{dx}{\sqrt[3]{3x+4}}}$ .

Formula used:

  ${\displaystyle \int x^{n}dx}=\frac{x^{n+1}}{n+1}+C,n\ne -1$ .

Calculation:

Consider the given integral,

  ${\displaystyle \int \frac{dx}{\sqrt[3]{3x+4}}}$

Now put $u=3x+4$ . Then,

  $\begin{array}{c}\frac{du}{dx}=3\left(1\right)+0\\ \frac{du}{dx}=3\\ \frac{du}{3}=dx\end{array}$

Therefore the integral becomes,

  $\begin{array}{c}{\displaystyle \int \frac{dx}{\sqrt[3]{3x+4}}}={\displaystyle \int \frac{\raisebox{1ex}{$du$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\sqrt[3]{u}}}\\ =\frac{1}{3}{\displaystyle \int \frac{du}{u^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}\\ =\frac{1}{3}{\displaystyle \int u^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}du}\\ =\frac{1}{3}\cdot \frac{u^{-\frac{1}{3}+1}}{{}^{-\frac{1}{3}+1}}+C\\ =\frac{1}{3}\cdot \frac{u^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{{}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+C\\ =\frac{1}{2}{\left(3x+4\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+C\end{array}$