MATH 152 Chapter 8.1

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Integration by Parts

$\mathrm {d} (uv)=u\,\mathrm {d} v+v\,\mathrm {d} u$

$\int {u\,\mathrm {d} v}=uv-\int {v\,\mathrm {d} u}$

Example

$\int {x^{2}e^{-3x}\,\mathrm {d} x}=?$

$u=x^{2}\,\!$	$\mathrm {d} v=e^{-3x}\,\mathrm {d} x$
$\mathrm {d} u=2x\,\mathrm {d} x$	$v=-{\frac {1}{3}}e^{-3x}$

$\int {x^{2}e^{-3x}\,\mathrm {d} x}\quad =\quad (x^{2})(-{\frac {1}{3}}e^{-3x})\quad +\quad \int {({\frac {1}{3}}e^{-3x})(2x\,\mathrm {d} x)}$

(Integrate the second integral by parts again)

$u=x\,\!$	$\mathrm {d} v=e^{-3x}\,\mathrm {d} x$
$\mathrm {d} u=\mathrm {d} x$	$v=-{\frac {1}{3}}e^{-3x}$

$\int {x^{2}e^{-3x}\,\mathrm {d} x}\quad =\quad (x^{2})(-{\frac {1}{3}}e^{-3x})\quad +\quad \int {({\frac {1}{3}}e^{-3x})(2x\,\mathrm {d} x)}$ ...

$=-{\frac {1}{3}}x^{2}e^{-3x}+{\frac {2}{3}}(-{\frac {1}{3}}xe^{-3x}-{\frac {1}{9}}e^{-3x})+C$

Integration by Tabular Method

DOES NOT ALWAYS WORK! $\mathrm {d} v$ must be have recursive integration

$u$	$\mathrm {d} v$
$x^{2}$	$e^{-3x}$
$2x$	$-{\frac {1}{3}}e^{-3x}$
$2$	${\frac {1}{9}}e^{-3x}$
$0$	$-{\frac {1}{27}}e^{-3x}$

(integrate until $u=0$, and integrate $\mathrm {d} v$ that many times.)

Multiply first $u$ by second $\mathrm {d}$, term is positive. Multiply second $u$ by third $\mathrm {d} v$, term is negative. Multiply third $u$ by fourth $\mathrm {d} v$, term is positive. (and so on)

Example 2

$\int _{0}^{1}{{\frac {x^{3}}{\sqrt {x^{2}+1}}}\,\mathrm {d} x}$

$u=x^{2}\,\!$	$\mathrm {d} v={\frac {x}{\sqrt {x^{2}+1}}}\,\mathrm {d} x$
$\mathrm {d} u=2x\,\mathrm {d} x$	$v={\sqrt {x^{2}+1}}$

$=x^{2}{\sqrt {x^{2}+1}}\,{\bigg |}_{0}^{1}-\int _{0}^{1}{2x\,{\frac {x}{\sqrt {x^{2}+1}}}\,\mathrm {d} x}$