Integration by parts
Some functions are a product of two very different pieces: a polynomial times an exponential, a "lone" logarithm, a sine times an $x$. None of the formulas in the table of standard integrals solves them directly. Integrating by parts is, quite literally, reading the derivative of a product backwards: it turns one integral into another (easier) one minus a term that is already solved.
Choosing $u$ and $dv$: the ALPES rule
The mnemonic ALPES (in Spanish) gives the order for choosing $u$: Arcs (arcsine, arctangent), Logarithms, Polynomials, Exponentials, Sines/cosines. The earlier a type of function appears in that list, the sooner it should be chosen as $u$; whatever is left is $dv$.
To get started with the integration by parts method, it is essential to know how to choose which function will be $u$ (the one we differentiate) and which will be $dv$…
To get started with the integration by parts method, it is essential to know how to choose which function will be $u$ (the one we differentiate) and which will be $dv$ (the one we integrate). Using the mnemonic rule ALPES (Arcs, Logarithms, Polynomials, Exponentials, Sines/Cosines), identify $u$ and $dv$ to solve the following indefinite integral, applying the formula $\int u\,dv = u\cdot v - \int v\,du$:
$$\int x \cdot e^{2x} \, dx$$Find the antiderivative of the function $f(x) = x \cdot \cos(x)$. Remember that when applying the integration by parts method, you should aim for the new resulting…
Find the antiderivative of the function $f(x) = x \cdot \cos(x)$. Remember that when applying the integration by parts method, you should aim for the new resulting integral to be simpler than the original one.
The logarithm: when $dv = dx$
The logarithm has no standard antiderivative and, according to ALPES, it is always chosen as $u$. If it is on its own (not multiplied by anything other than a constant), $dv$ is simply $dx$.
A classic integration by parts example is the logarithmic function, as it does not have a direct standard integral. In this case, we must consider that the function is…
A classic integration by parts example is the logarithmic function, as it does not have a direct standard integral. In this case, we must consider that the function is multiplied by $1$ (that is, $dv = dx$).
Calculate the following indefinite integral:
$$\int \ln(x) \, dx$$Calculate the indefinite integral of the function $f(x) = \ln(3x + 2)$ using the integration by parts method. Hint: you can consider that the function is multiplied by 1…
Calculate the indefinite integral of the function $f(x) = \ln(3x + 2)$ using the integration by parts method. Hint: you can consider that the function is multiplied by 1 to choose your variables $u$ and $dv$.
Integrating by parts twice, and cyclic integrals
If the polynomial has degree $2$ or higher, a single application of integration by parts is not enough: each time you integrate by parts, the degree of the polynomial drops by one, so the process has to be repeated.
There is a special case in which, after integrating by parts twice, the original integral reappears on the right-hand side of the equation. It is then treated as an equation: you solve for the integral you are looking for.
The marginal profit of a vintage clothing shop (in hundreds of euros per month) is defined by the function $B'(x) = x^2 \cdot e^{-x}$, where $x$ represents the number of…
The marginal profit of a vintage clothing shop (in hundreds of euros per month) is defined by the function $B'(x) = x^2 \cdot e^{-x}$, where $x$ represents the number of months since its opening. Applying the integration by parts method as many times as necessary, find the expression for the total profit $B(x)$, knowing that at the end of the first month ($x = 1$) the profit was zero.
An engineer models the vibration velocity of a structure subjected to forced oscillations using the function (in metres per second): $$v(t) = e^{t} \cdot \cos(t)$$…
An engineer models the vibration velocity of a structure subjected to forced oscillations using the function (in metres per second):
$$v(t) = e^{t} \cdot \cos(t)$$Calculate the indefinite integral of this function. Note that this type of integral is called "cyclical", since by applying the method of integration by parts twice, the original integral will reappear on the right-hand side of the equality.
Applications: areas and definite integrals
The procedure with definite integrals is the usual one: you find the antiderivative by parts and apply the Fundamental Theorem of Calculus (Barrow's rule) at the end, evaluating at both limits.
Determine the definite integral over the interval $[1, e]$ of the function $f(x) = 5 \ln(x)$.
Determine the definite integral over the interval $[1, e]$ of the function $f(x) = 5 \ln(x)$.
Find the area of the region bounded by the graph of the function $f(x) = (x - 1) \cdot e^x$, the $OX$-axis and the lines $x = 0$ and $x = 1$. Draw a small sketch of the…
Find the area of the region bounded by the graph of the function $f(x) = (x - 1) \cdot e^x$, the $OX$-axis and the lines $x = 0$ and $x = 1$. Draw a small sketch of the region before calculating the integral.