next up previous
Next: Logarithmic integration Up: Methods of integration Previous: Integration by substitution

Integration by parts

This is the integral calculus counterpart of the product rule in differential calculus. To derive it, we integrate the expression

\begin{displaymath}
\frac{\mathrm{d}}{\mathrm{d}x} (uv) = v \frac{\mathrm{d}u}{\mathrm{d}x} + u \frac{\mathrm{d}v}{\mathrm{d}x}
\end{displaymath}

to obtain

\begin{eqnarray*}
\int \frac{\mathrm{d}}{\mathrm{d}x} (uv) \mathrm{d}x & = & \in...
...t u \mathrm{d}v \\
\int u \mathrm{d}v = uv - \int v \mathrm{d}u
\end{eqnarray*}



For definite integrals,

\begin{displaymath}
\int_{a}^{b} u \mathrm{d}v = \Biggl [ uv - \int v \mathrm{d}...
...b} = \Biggl [ uv \Biggr
]_{a}^{b} - \int_{a}^{b} v \mathrm{d}u
\end{displaymath}

We can find \( \int x \ln{x} \mathrm{d}x \) as follows.

\begin{eqnarray*}
\int x \ln{x} \mathrm{d}x & = & \int \ln{x} \mathrm{d}(\frac{x...
...\mathrm{d}x \\
& = & \int x \mathrm{d}x \\
& = & \frac{x^2}{2}
\end{eqnarray*}



so that

\begin{displaymath}
\int x \ln{x} \mathrm{d}x = \frac{1}{2} x^2 \ln{x} - \frac{1}{4} x + c
\end{displaymath}

We can also use integration by parts to find integrals which can not be evaluated directly such as \( \int \ln{x} \mathrm{d}x \). This is done as follows.

\begin{eqnarray*}
\int \ln{x} \mathrm{d}x & = & x \ln{x} - \int x \mathrm{d}(\ln...
...thrm{d}x \\
& = & x \ln{x} - x + c \\
& = & x (\ln{x} - 1) + c
\end{eqnarray*}





Alexander Frolkin 2001-03-13