Integration by parts

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

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

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,

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

We can find 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

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

We can also use integration by parts to find integrals which can not be evaluated directly such as . This is done as follows.

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