MHB S6.7.1.13 natural log Integration

[karush]

$\large {S6.7.1.13}$
$\tiny\text {natural log Integration}$
$$\displaystyle
\int e^{\sqrt[3]{x}} \, dx
= 3\left(x^\frac{2}{3}
-2\sqrt[3]{x}
+2\right){e}^\sqrt[3]{x}+C \\
u=x^{1/3} \therefore 3{x}^{\frac{2}{3}} du
= dx $$
$\text{not sure if this is how to start to get to a 3 term answer} $
$\tiny\text{ Surf the Nations math study group}$

 

[Greg]

You're on the right track...

$$u=x^{1/3},\quad u^3=x,\quad du=\dfrac13x^{-2/3}\Rightarrow3u^2du=dx$$

Now make the sub and use integration by parts twice.

 

[karush]

$\large {S6.7.1.13}$
$\tiny\text {natural log Integration}$
$$\displaystyle
\int e^{\sqrt[3]{x}} \, dx
= 3\left(x^\frac{2}{3}
-2\sqrt[3]{x}
+2\right){e}^\sqrt[3]{x}+C \\
u=x^{1/3} \therefore 3{u}^{2} du
= dx $$
$\text{so then we have} $
$$\displaystyle
3\int{u}^{2}e^{u} \ dx$$
$\text{looks like IBP 2 times} $

$\tiny\text{ Surf the Nations math study group}$

 

[karush]

$\large {S6.7.1.13}$
$\tiny\text {natural log Integration}$
$$\displaystyle
I= \int e^{\sqrt[3]{x}} \, dx \\
u=x^{1/3} \therefore 3{u}^{2} du= dx $$
$\text{IBP 1}$
$$\displaystyle
I=3\int{u}^{2}e^{u} \ du \\
\begin{align}
{u_1} & = {u}^{2} & {dv_1}&={e}^{u} \ d{u} & \\
\frac{1}{2}{du_1}&=du& {v_1}&={e}^{u}
\end{align}$$
$\text{IBP 2}$
$$\displaystyle
I=3\left[{u}^{2}{e}^{u}-2\int u e^{u} \ du \right] \\

\begin{align}
{u_2} & = {u}^{2} & {dv_2 }&={e}^{u} \ d{u} & \\
{du_2}&=du& {v_2}&={e}^{u}
\end{align}$$
$\text{factor and back substittute}$
$$I= 3\left(x^\frac{2}{3}
-2\sqrt[3]{x}
+2\right){e}^\sqrt[3]{x}+C
$$
$\tiny\text{ Surf the Nations math study group}$