Pauls Online Notes : Calculus II - Integration by Parts
Click the link and scroll down.
Look at Example 7. In this example why does v = 2/9 ... I do not see were the 2/9 comes from. I also do not see where the -4/45 comes from.
Pauls Online Notes : Calculus II - Integration by Parts
Click the link and scroll down.
Look at Example 7. In this example why does v = 2/9 ... I do not see were the 2/9 comes from. I also do not see where the -4/45 comes from.
If $\displaystyle
dv = x^2 \sqrt {x^3 + 1} \Rightarrow \int {(dv} )dx = v
$
Then:
$\displaystyle
\begin{gathered}
\int {x^2 \sqrt {x^3 + 1} } dx \hfill \\
u^2 = x^3 + 1 \Rightarrow 2udu = 3x^2 dx \Leftrightarrow dx = \frac{{2u}}
{{3x^2 }}du \hfill \\
\int {\frac{2}
{3}\sqrt u \cdot udu = \frac{2}
{3}\int {u^2 du} = \frac{2}
{3} \cdot \frac{{u^3 }}
{3}} = \frac{2}
{9} \cdot \left( {u^2 } \right)^{3/2} = \frac{2}
{9} \cdot \left( {x^3 + 1} \right)^{3/2} = v \hfill \\
\end{gathered}
$
I donīt understand why use integration by parts, is more hard than use the subtitution $\displaystyle u^2=x^3+1$