Methods of Integration

(cod: P-71-62-1) Integration by Parts: This is one of the main techniques of integration, being a consequence of the product rule. Indeed, let \(f\) and \(g\) be two functions with continuous derivatives on an open interval \(I\). By the Product Rule, we have \[\dfrac{d}{dx} f(x)g(x) = f'(x)g(x) + f(x)g'(x),\] or, equivalently, \[\small{f(x)g'(x) = \dfrac{d}{dx} \left(f(x)g(x)\right) - g(x)f'(x).} \ \ \ \ \ (1)\] Taking the antiderivative on both sides, we obtain \[\small{\int f(x)g'(x)\, dx = f(x)g(x) - \int g(x)f'(x)\, dx},\] where we omit the constant of integration when integrating the first term on the right-hand side, as this constant will be introduced when solving the integral. The technique consists of replacing the integral on the left-hand side with the expression on the right-hand side. Taking \(u = f(x)\) and \(dv = g'(x)dx\) in the integral on the left-hand side and considering that \(du = f'(x)dx\) and \(v = g(x)\), we can rewrite the above formula in a more compact form: \[\int u \, dv = uv - \int v \, du.\] When calculating a definite integral over an interval \([a,b] \subset I\), we can integrate both sides of equation (1), obtaining \begin{eqnarray*} & & \int_a^b f(x)g'(x)\, dx \\ & & \\ &=& f(x)g(x)\Big{]}_a^b + \int_a^b g(x)f'(x)\, dx. \end{eqnarray*} Consider the questions below:

(1) Suppose we want to calculate the integral \[\int x e^x \, dx\] using integration by parts and, for this, we set \(u = x\) and \(dv = e^x dx\). What will be our conclusion?

(2) Suppose we want to calculate the integral \[\int_1^e \ln{x} \, dx\] using integration by parts and, for this, we choose \(u = \ln{x}\) and \(dv = dx\) (or, equivalently, \(dv/dx = 1\)). What will we obtain?

(1) \begin{eqnarray*} \int x e^x \, dx &=& xe^x - \int e^x \, dx \\ & & \\ &=& e^x(x-1) + C. \end{eqnarray*}
(2) \begin{eqnarray*} \int_1^e \ln{x}\, dx &=& x \Big{]}_1^e - \int_1^e x \cdot \dfrac{1}{x} \, dx \\ & & \\ &=& e-1 - (e-1) = 0. \end{eqnarray*}

(1) \begin{eqnarray*} \int x e^x \, dx &=& e^x - \int e^x \, dx \\ & & \\ &=& 0 + C = C. \end{eqnarray*}
(2) \begin{eqnarray*} \int_1^e \ln{x}\, dx &=& x \ln{x}\Big{]}_1^e - \int_1^e x \cdot \dfrac{1}{x} \, dx \\ & & \\ &=& e - (e-1) = 1. \end{eqnarray*}

(1) \begin{eqnarray*} \int x e^x \, dx &=& xe^x - \int e^x \, dx \\ & & \\ &=& e^x(x-1) + C. \end{eqnarray*}
(2) \begin{eqnarray*} \int_1^e \ln{x}\, dx &=& x \ln{x}\Big{]}_1^e - \int_1^e x \cdot \dfrac{1}{x} \, dx \\ & & \\ &=& e - (e-1) = 1. \end{eqnarray*}

(1) \begin{eqnarray*} \int x e^x \, dx &=& xe^x - \int e^x \, dx \\ & & \\ &=& e^x(x-1) + C. \end{eqnarray*}
(2) \begin{eqnarray*} \int_1^e \ln{x}\, dx &=& \ln{x}\Big{]}_1^e - \int_1^e x \cdot \dfrac{1}{x} \, dx \\ & & \\ &=& 1 - (e-1) = 2-e. \end{eqnarray*}

(1) \begin{eqnarray*} \int x e^x \, dx &=& x- \int e^x \, dx \\ & & \\ &=& x - e^x + C. \end{eqnarray*}
(2) \begin{eqnarray*} \int_1^e \ln{x}\, dx &=& \ln{x}\Big{]}_1^e - \int_1^e x \cdot \dfrac{1}{x} \, dx \\ & & \\ &=& 1 - (e-1) = 2-e. \end{eqnarray*}