Editing Integration by parts (section)

====Functions multiplied by unity====

Two other well-known examples are when integration by parts is applied to a function expressed as a product of 1 and itself. This works if the derivative of the function is known, and the integral of this derivative times <math>x</math> is also known.

The first example is <math>\int \ln(x) dx</math>. We write this as:

<math display="block">I=\int\ln(x)\cdot 1\,dx\,.</math>

Let:

<math display="block">u = \ln(x)\ \Rightarrow\ du = \frac{dx}{x}</math>
<math display="block">dv = dx\ \Rightarrow\ v = x</math>

then:

<math display="block">
\begin{align}
\int \ln(x)\,dx & = x\ln(x) - \int\frac{x}{x}\,dx \\
& = x\ln(x) - \int 1\,dx \\
& = x\ln(x) - x + C
\end{align}
</math>

where <math>C</math> is the [[constant of integration]].

The second example is the [[inverse tangent]] function <math>\arctan(x)</math>:

<math display="block">I=\int\arctan(x)\,dx.</math>

Rewrite this as

<math display="block">\int\arctan(x)\cdot 1\,dx.</math>

Now let:

<math display="block">u = \arctan(x)\ \Rightarrow\ du = \frac{dx}{1+x^2}</math>

<math display="block">dv = dx\ \Rightarrow\ v = x</math>

then

<math display="block">
\begin{align}
\int\arctan(x)\,dx
& = x\arctan(x) - \int\frac{x}{1+x^2}\,dx \\[8pt]
& = x\arctan(x) - \frac{\ln(1+x^2)}{2} + C
\end{align}
</math>

using a combination of the [[inverse chain rule method]] and the [[natural logarithm integral condition]].