Integration by Parts

Failure is not an option. Failure is necessary…We try. We fail. We learn. We adjust, Craig Groeschel.

Core Calculus Concepts: The Antiderivative

In calculus, an antiderivative (or indefinite integral) of a function $f(x)$ is a function $G(x)$ that can be differentiated to obtain the original function. In mathematical terms, if $G'(x) = f(x)$, then, $G(x) = \int f(x)dx$

This operation is the fundamental inverse of differentiation.

There are several techniques used for integration, each suited to different types of integrals. The most common methods include U-Substitution (or change of variables), Integration by Parts, Trigonometric Substitution, Partial Fraction Decomposition, Improper Integrals, and the use of special functions.

Before diving into Integration by Parts, let’s briefly touch on the first major tool in your toolkit: U-Substitution. The general form of an integrand that requires U-Substitution is $\int f(g(x))g'(x) \, dx$. This can be cleanly rewritten as $\int f(u) \, du$. A big hint to use U-Substitution is when you spot a composition of functions where the derivative of the inner function is also present in the integrand (essentially, it’s the reverse chain rule).

Integration by parts

Integration by parts is a powerful technique used primarily to integrate the product of two functions. Just as U-Substitution is the reverse of the chain rule, Integration by Parts is the reverse of the product rule.

The product rule for derivatives states that the derivative of a product of two functions is the first times the derivative of the second, plus the second times the derivative of the first: $(uv)' = u'v + uv'$

If we rearrange this to isolate $uv'$ ($uv' = (uv)' - u'v$) and integrate both sides with respect to $x$, we get the fundamental formula for integration by parts: $\boxed{\int u \, dv = uv - \int v \, du}$, e.g., $\int \ln(x) dx = [u = \ln(x), u' = \frac{1}{x}dx, v = x, v' = dx] x\ln(x) -\int \frac{1}{x} x dx = x\ln(x) -x + C.$

The LIATE Strategy

A popular strategy to select which function should be $u$ (the one to differentiate) and which should be $dv$ (the one to integrate) is the LIATE mnemonic:

1. Logarithmic ($\ln(x)$, $\log_2(x)$)
2. Inverse trigonometric ($\arcsin(x)$, $\arctan(x)$)
3. Algebraic ($x^n$, $x^2+4$, $3x+1$)
4. Trigonometric ($\sin(x)$, $\cos(x)$, $\tan(x)$)
5. Exponential ($e^x$, $2^x$)

How it works: The function that appears first in this list is usually the best choice for $u$ (because it simplifies nicely when differentiated). The function that appears last in the list is the best choice for $dv$ (because it is easily integrated).

The moral of the story is that mnemonics are useful guides, but they aren’t set in stone. Nothing replaces your mathematical intuition and experience.

Geometrical interpretation

The integration by parts formula isn’t just algebraic magic —it has a beautiful visual proof based on calculating areas.

Let’s look at a curve on a coordinate plane defined by a function $y = f(x)$. Suppose we want to find the relationship between the areas associated with points $(x_1, y_1)$ and $(x_2, y_2)$.

If we draw a large rectangle from the origin to $(x_2, y_2)$ and subtract the smaller rectangle from the origin to $(x_1, y_1)$, the total remaining L-shaped area is: $\text{Total Area} = x_2 y_2 - x_1 y_1$

Notice that this L-shaped region can be split into two distinct areas:

1. $A_1$, Area under the curve: Integrating with respect to the x-axis. This is $\int_{x_1}^{x_2} y \, dx$.
2. $A_2$, Area between the curve and the y-axis: If we swap our perspective and treat $x$ as a function of $y$, this area is $\int_{y_1}^{y_2} x \, dy$.

Because these two areas perfectly combine to form the L-shape, we get: $\int_{x_1}^{x_2} y \, dx + \int_{y_1}^{y_2} x \, dy = x_2 y_2 - x_1 y_1$

If we define the curve parametrically where $x = u(t)$ and $y = v(t)$, the geometric area equation perfectly matches the integration by parts formula: $\int u \, dv + \int v \, du = uv \bigg\vert{}_{a}^{b}$

Rearranging this gives us our familiar formula: $\int u \, dv = uv \bigg\vert{}_{a}^{b} - \int v \, du$

Solved exercises

• U-Substitution. Evaluate: $\int 2x^2\sqrt{x+1} \, dx$

Let $u = x + 1$, which means $du = dx$.

Because we have an $x^2$ term outside the square root, we also need to solve for $x$: $x = u - 1 \implies x^2 = (u-1)^2 = u^2 - 2u + 1$

Now, substitute everything into the integral: $\int 2x^2\sqrt{x+1} \, dx = 2 \int (u-1)^2 \sqrt{u} \, du= 2 \int (u^2 - 2u + 1)u^{\frac{1}{2}} \, du = 2 \int \left( u^{\frac{5}{2}} - 2u^{\frac{3}{2}} + u^{\frac{1}{2}} \right) \, du$

Integrate term by term using the power rule: $= 2 \left( \frac{2}{7}u^{\frac{7}{2}} - 2\left(\frac{2}{5}\right)u^{\frac{5}{2}} + \frac{2}{3}u^{\frac{3}{2}} \right) + C$

Substitute $x + 1$ back in for $u$ to get the final answer: $\int 2x^2\sqrt{x+1} \, dx = \frac{4}{7}(x+1)^{\frac{7}{2}} - \frac{8}{5}(x+1)^{\frac{5}{2}} + \frac{4}{3}(x+1)^{\frac{3}{2}} + C$

• Integration by Parts. $\int x \cos(x) \, dx$

Using the LIATE rule, Algebraic ($x$) comes before Trigonometric ($\cos(x)$), so we assign our variables as follows: $u = x \implies du = dx$, $dv = \cos(x) \, dx \implies v = \sin(x)$

Apply the formula $\int u \, dv = uv - \int v \, du$: $\int x \cos(x) \, dx = x\sin(x) - \int \sin(x) \, dx = x\sin(x) - (-\cos(x)) + C= x\sin(x) + \cos(x) + C$

• U-Substitution. $\int 3x^2(x^3+5)^7 \, dx$

This is a classic U-substitution problem because we have a function $(x^3+5)$ and its exact derivative ($3x^2$) sitting outside of it.

Let $u = x^3 + 5$. Then, $du = 3x^2 \, dx$

Substitute $u$ and $du$ into the integral: $\int u^7 \, dx = \frac{u^8}{8} + C$

Substitute $x$ back in our original expression: $\int 3x^2(x^3+5)^7 \, dx = \frac{(x^3+5)^8}{8} + C$

• Standard Integration by Parts. $\int 6x^2 \ln(x) \, dx$

Using the LIATE rule, Logarithmic ($\ln(x)$) comes before Algebraic ($6x^2$). Let $u = \ln(x) \implies du = \frac{1}{x} \, dx, dv = 6x^2 \, dx \implies v = 2x^3$

Applying $\int u \, dv = uv - \int v \, du$: $\int 6x^2 \ln(x) \, dx = 2x^3 \ln(x) - \int 2x^3 \left(\frac{1}{x}\right) \, dx = 2x^3 \ln(x) - \int 2x^2 \, dx = 2x^3 \ln(x) - \frac{2}{3}x^3 + C$

• Integration by Parts with a Sub-Substitution. $\int 2x(3x-2)^6 \, dx$

Let $u = 2x \implies du = 2 \, dx$ and $dv = (3x-2)^6 \, dx$

To find $v$, we need a quick U-substitution: let $w = 3x-2 \implies dw = 3 \, dx$.

$v = \int (3x-2)^6 \, dx = \int \frac{1}{3}w^6 \, dw = \frac{1}{21}w^7 = \frac{1}{21}(3x-2)^7$

Now, apply the parts formula: $= 2x \left[ \frac{1}{21}(3x-2)^7 \right] - \int \frac{2}{21}(3x-2)^7 \, dx$

For the remaining integral, use the same U-substitution ($w = 3x-2$): $- \int \frac{2}{21}(3x-2)^7 \, dx = -\frac{2}{21} \int \frac{1}{3} w^7 \, dw = -\frac{2}{63} \left( \frac{w^8}{8} \right) = -\frac{2}{504}(3x-2)^8 = -\frac{1}{252}(3x-2)^8$

Combine and factor for the final answer: $\int 2x(3x-2)^6 \, dx = \frac{2x}{21}(3x-2)^7 - \frac{1}{252}(3x-2)^8 + C = \frac{24x}{252}(3x-2)^7 - \frac{3x-2}{252}(3x-2)^7 + C = \frac{21x+2}{252}(3x-2)^7 + C$

• Repeated Integration by Parts. $\int x^2\sin(x) \, dx$

Because we have an $x^2$ term, we will need to perform integration by parts twice to lower the power to zero.

1. First Pass: $u = x^2 \implies du = 2x \, dx, dv = \sin(x) \, dx \implies v = -\cos(x)$
   $\int x^2\sin(x) \, dx = -x^2\cos(x) - \int -\cos(x)(2x) \, dx = -x^2\cos(x) + 2\int x\cos(x) \, dx$
2. Second Pass (for $\int x\cos(x) \, dx$): $u = x \implies du = dx$, $dv = \cos(x) \, dx \implies v = \sin(x)$
   $2\int x\cos(x) \, dx = 2 \left( x\sin(x) - \int \sin(x) \, dx \right) = 2(x\sin(x) - (-\cos(x))) = 2x\sin(x) + 2\cos(x)$
3. Final Assembly: $\int x^2\sin(x) \, dx = -x^2\cos(x) + 2x\sin(x) + 2\cos(x) + C$

Reduction Formulas

Reduction formulas use integration by parts to establish a pattern, reducing the power of $n$ by $1$ each time.

• Logarithmic Reduction.

Base case: $\int (\ln x)^2 \, dx$

$u = (\ln x)^2 \implies du = 2\ln(x)\left(\frac{1}{x}\right) \, dx, dv = dx \implies v = x$

$\int (\ln x)^2 \, dx = x(\ln x)^2 - \int 2\ln(x)\left(\frac{1}{x}\right)x \, dx = x(\ln x)^2 - 2\int \ln(x) \, dx$

Since we know $\int \ln(x) \, dx = x\ln(x) - x$, this becomes: $= x(\ln x)^2 - 2(x\ln(x) - x) + C$

General Case: $\int (\ln x)^n \, dx$

$u = (\ln x)^n \implies du = n(\ln(x))^{n-1}\left(\frac{1}{x}\right) \, dx, dv = dx \implies v = x$

Using the exact same logic, we establish the recurrence relation $F_n(x)$: $F_n(x) = \int (\ln x)^n \, dx = x(\ln x)^n - n\int (\ln x)^{n-1} \, dx$

$\boxed{F_n(x) = x(\ln x)^n - nF_{n-1}(x)}$

• Exponential Reduction

General Case: $\int x^n e^x \, dx$

$u = x^n \implies du = nx^{n-1} \, dx, dv = e^x \, dx \implies v = e^x$

$F_n(x) = x^n e^x - \int nx^{n-1}e^x \, dx$. Therefore, $\boxed{F_n(x) = x^n e^x - nF_{n-1}(x)}$

• Splitting the Integrand. $\int x^5\sqrt{x^3+1} \, dx$

This is a clever problem. A standard U-substitution fails initially because the derivative of $x^3+1$ is $3x^2$, but we have $x^5$. The trick is to algebraically split $x^5$ into $x^3 \cdot x^2$.

By doing this, we can set up an integration by parts where the $dv$ component perfectly accepts a U-substitution!

Let $u = x^3 \implies du = 3x^2 \, dx, dv = x^2\sqrt{x^3+1} \, dx$

To find $v$, use substitution ($r = x^3+1 \implies dr = 3x^2 \, dx$): $v = \int x^2\sqrt{x^3+1} \, dx = \frac{1}{3}\int \sqrt{r} \, dr = \frac{1}{3}\left(\frac{2}{3}r^{\frac{3}{2}}\right) = \frac{2}{9}(x^3+1)^{\frac{3}{2}}$

Now apply the parts formula: $= x^3\left[ \frac{2}{9}(x^3+1)^{\frac{3}{2}} \right] - \int \frac{2}{9}(x^3+1)^{\frac{3}{2}}(3x^2) \, dx = \frac{2}{9}x^3(x^3+1)^{\frac{3}{2}} - \frac{2}{3}\int x^2(x^3+1)^{\frac{3}{2}} \, dx$

Use the exact same substitution for this final integral ($r = x^3+1$): $= \frac{2}{9}x^3(x^3+1)^{\frac{3}{2}} - \frac{2}{3} \left[ \frac{1}{3} \cdot \frac{2}{5} (x^3+1)^{\frac{5}{2}} \right] = \frac{2}{9}x^3(x^3+1)^{\frac{3}{2}} - \frac{4}{45}(x^3+1)^{\frac{5}{2}} + C$

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