structural:summarization
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 x {\displaystyle x} is also known.

The first example is ∫ ln (x) d x {\displaystyle \int \ln(x)dx}. We write this as:

I = ∫ ln (x) ⋅ 1 d x. {\displaystyle I=\int \ln(x)\cdot 1\,dx\,.}

Let:

u = ln (x) ⇒ d u = d x x {\displaystyle u=\ln(x)\ \Rightarrow \ du={\frac {dx}{x}}} d v = d x ⇒ v = x {\displaystyle dv=dx\ \Rightarrow \ v=x}

then:

∫ ln (x) d x = x ln (x) − ∫ x x d x = x ln (x) − ∫ 1 d x = x ln (x) − x + C {\displaystyle {\begin{aligned}\int \ln(x)\,dx&=x\ln(x)-\int {\frac {x}{x}}\,dx\\&=x\ln(x)-\int 1\,dx\\&=x\ln(x)-x+C\end{aligned}}}

where C {\displaystyle C} is the constant of integration.

The second example is the inverse tangent function arctan (x) {\displaystyle \arctan(x)}:

I = ∫ arctan (x) d x. {\displaystyle I=\int \arctan(x)\,dx.}

Rewrite this as

∫ arctan (x) ⋅ 1 d x. {\displaystyle \int \arctan(x)\cdot 1\,dx.}

Now let:

u = arctan (x) ⇒ d u = d x 1 + x 2 {\displaystyle u=\arctan(x)\ \Rightarrow \ du={\frac {dx}{1+x^{2}}}}

d v = d x ⇒ v = x {\displaystyle dv=dx\ \Rightarrow \ v=x}

then

∫ arctan (x) d x = x arctan (x) − ∫ x 1 + x 2 d x = x arctan (x) − ln (1 + x 2) 2 + C {\displaystyle {\begin{aligned}\int \arctan(x)\,dx&=x\arctan(x)-\int {\frac {x}{1+x^{2}}}\,dx\\&=x\arctan(x)-{\frac {\ln(1+x^{2})}{2}}+C\end{aligned}}}

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