Integrals · Lesson 6 of 9

Integration by Parts and Radical Forms

“Reverse the product rule, then use it to derive integrals containing logarithms, inverse functions, and square roots.”

Learning Objectives

• Derive integration by parts from the product rule and choose useful factors. • Use repeated integration by parts, including an integral that reappears. • Recognise and justify the exponential product-derivative pattern. • Derive the three standard quadratic square-root integrals. • Complete the square to integrate shifted radical expressions.

Reading the product rule backwards

Substitution reverses the chain rule. Integration by parts reverses the product rule and is especially useful when differentiating one factor makes it simpler. It replaces one product integral by another; the method helps only if the new integral is easier to handle or can be related back to the original one.

Start from $(uv)' = uv' + v u'$ . Integrating both sides gives $uv = \int uv' d x + \int v u' d x$ , with a constant ultimately included in the final answer. Rearranging gives the integration-by-parts formula.

Integration by partsLaTeX

\int u d v = uv - \int v d u

Here

u = u (x)

d u = u' (x) d x

d v = v' (x) d x

, and

v

is one antiderivative of the other factor. Choose

u

so that its derivative is helpful and

d v

so that its integral is available.

In the first-function and second-function language, the result is: first function multiplied by the integral of the second, minus the integral of the derivative of the first multiplied by the integral of the second. One factor may be the constant function $1$ , so the method also applies to a lone logarithm or inverse trigonometric function.

Example — Choosing the useful first function

Problem
Evaluate $\int x cos x d x$ .

1.Take $u = x$ and $d v = cos x d x$ . Then $d u = d x$ and $v = sin x$ .
2.The formula gives $x sin x - \int sin x d x = x sin x + cos x + C$ .
3.If instead $u = cos x$ and $d v = x d x$ , the remainder is $\frac{1}{2} \int x^{2} sin x d x$ , which is more complicated. The formula is still true, but the choice is unhelpful.

When one factor is a polynomial and the other has a simple integral, differentiating the polynomial often reduces its degree. Logarithmic and inverse trigonometric functions are often useful choices for $u$ because their derivatives are algebraic. These are choices guided by the resulting integral, rather than an inflexible rule based on the order of symbols.

Why no extra constant is needed for the inner antiderivative

If $v$ is replaced by $v + k$ , then $u (v + k) - \int (v + k) d u = uv + k u - \int v d u - k u + C$ . The terms involving $k$ cancel. Use one convenient antiderivative for $v$ and put a single arbitrary constant in the final result.

Logarithms, exponentials, and inverse functions

The first examples show how the method converts unfamiliar antiderivatives into familiar ones. A single application is sometimes enough. In other cases the same choice can be repeated until the difficult factor has been differentiated away.

Example — Integrating a logarithm

Problem
Evaluate $\int ln x d x$ for $x > 0$ .

1.View the integrand as $(ln x) \cdot 1$ . Choose $u = ln x$ and $d v = d x$ .
2.Then $d u = d x / x$ and $v = x$ , so the result is $x ln x - \int x (1/ x) d x$ .
3.Thus $\int ln x d x = x ln x - x + C$ . Differentiation produces $ln x + 1 - 1 = ln x$ .

Example — A polynomial times an exponential

Problem
Evaluate $\int x e^{x} d x$ and $\int x^{2} e^{x} d x$ .

1.For the first choose $u = x$ , $d v = e^{x} d x$ . The result is $x e^{x} - \int e^{x} d x = e^{x} (x - 1) + C$ .
2.For the second choose $u = x^{2}$ and the same $d v$ . This gives $x^{2} e^{x} - 2 \int x e^{x} d x$ .
3.Substitute the first result to obtain $e^{x} (x^{2} - 2 x + 2) + C$ . The polynomial degree decreases at each use of the formula.

Example — An inverse sine multiplied by its helpful partner

Problem
Evaluate $\int x sin^{- 1} x / 1 - x^{2} d x$ on $- 1 < x < 1$ .

1.Take $u = sin^{- 1} x$ and $d v = x d x / 1 - x^{2}$ . The second factor integrates using $t = 1 - x^{2}$ , giving $v = - 1 - x^{2}$ .
2.Since $d u = d x / 1 - x^{2}$ , integration by parts gives $- 1 - x^{2} sin^{- 1} x + \int 1 d x$ .
3.The answer is $x - 1 - x^{2} sin^{- 1} x + C$ . Alternatively, putting $x = sin θ$ first leads to a product integral in $θ$ .

Example — A standalone inverse tangent

Problem
Evaluate $\int tan^{- 1} x d x$ .

1.Choose $u = tan^{- 1} x$ and $d v = d x$ , giving $d u = d x / (1 + x^{2})$ and $v = x$ .
2.The formula gives $x tan^{- 1} x - \int x / (1 + x^{2}) d x$ .
3.Set $t = 1 + x^{2}$ in the remainder. The final result is $x tan^{- 1} x - \frac{1}{2} ln (1 + x^{2}) + C$ .

When the original integral returns

Sometimes integration by parts cycles between two related integrals instead of making a factor disappear. This can still be useful. Treat the original integral as an unknown, collect its repeated occurrences, and solve the resulting algebraic equation.

Example — Two uses and one algebraic equation

Problem
Evaluate $I = \int e^{x} sin x d x$ .

1.Take $u = e^{x}$ and $d v = sin x d x$ . Then $I = - e^{x} cos x + J$ , where $J = \int e^{x} cos x d x$ .
2.Apply the method to $J$ with $u = e^{x}$ and $d v = cos x d x$ . This gives $J = e^{x} sin x - I$ .
3.Substitute into the first equation: $I = - e^{x} cos x + e^{x} sin x - I$ . Therefore $2 I = e^{x} (sin x - cos x)$ , up to an arbitrary constant.
4.Hence $I = \frac{1}{2} e^{x} (sin x - cos x) + C$ . Differentiating checks both the sign and the factor $1/2$ .

Recognising a complete exponential product derivative

Some expressions that appear to require integration by parts can be recognised immediately as a product derivative. Since differentiating $e^{x} f (x)$ gives $e^{x} [f (x) + f' (x)]$ , an integrand with exactly those two pieces can be integrated in one step. The recognition must be verified rather than guessed.

Exponential product patternLaTeX

\int e^{x} [f (x) + f' (x)] d x = e^{x} f (x) + C

The equality follows directly from the product rule. Integration by parts gives the same result because its remaining integral cancels the second term of the original integrand.

Example — Identify the function and its derivative

Problem
Evaluate $\int e^{x} [tan^{- 1} x + 1/ (1 + x^{2})] d x$ .

1.Set $f (x) = tan^{- 1} x$ . Its derivative is $1/ (1 + x^{2})$ , exactly the second term in brackets.
2.The integrand is therefore the derivative of $e^{x} tan^{- 1} x$ .
3.The answer is $e^{x} tan^{- 1} x + C$ .

Example — Reveal the same pattern algebraically

Problem
Evaluate $\int e^{x} (x^{2} + 1) / (x + 1)^{2} d x$ for $x \neq = - 1$ .

1.Write $(x^{2} + 1) / (x + 1)^{2} = (x - 1) / (x + 1) + 2/ (x + 1)^{2}$ . This follows because $x^{2} + 1 = (x - 1) (x + 1) + 2$ .
2.Take $f (x) = (x - 1) / (x + 1)$ . The quotient rule gives $f' (x) = 2/ (x + 1)^{2}$ .
3.Thus the answer is $e^{x} (x - 1) / (x + 1) + C$ . The algebraic rewrite has exposed both parts of a single product derivative.

Deriving integrals of quadratic square roots

We already know how to integrate the reciprocals of three quadratic square roots. Integration by parts now connects those results to the square roots themselves. Each derivation creates another copy of the original integral, so the earlier idea of collecting an unknown integral appears again.

Let $I = \int x^{2} - a^{2} d x$ , with $a > 0$ and $∣ x ∣ > a$ . Choose $u = x^{2} - a^{2}$ and $d v = d x$ . Since $d u = x d x / x^{2} - a^{2}$ and $v = x$ , the formula gives $I = x x^{2} - a^{2} - \int x^{2} / x^{2} - a^{2} d x$ .

Replace the numerator by $x^{2} = (x^{2} - a^{2}) + a^{2}$ . The remainder splits into $I + a^{2} \int d x / x^{2} - a^{2}$ . Thus $2 I = x x^{2} - a^{2} - a^{2} \int d x / x^{2} - a^{2}$ . Substitute the known reciprocal-root logarithm and divide by two.

Root of a difference of squaresLaTeX

\int x^{2} - a^{2} d x = \frac{x}{2} x^{2} - a^{2} - \frac{a ^{2}}{2} ln x + x^{2} - a^{2} + C

This is derived on intervals with

∣ x ∣ > a

. Endpoint values may be considered separately when a definite integral has a continuous root integrand.

For $J = \int x^{2} + a^{2} d x$ , use the same first step, but split $x^{2} = (x^{2} + a^{2}) - a^{2}$ . This gives $J = x x^{2} + a^{2} - J + a^{2} \int d x / x^{2} + a^{2}$ . Moving the repeated $J$ to the left explains why the logarithmic term now has a positive sign.

Root of a sum of squaresLaTeX

\int x^{2} + a^{2} d x = \frac{x}{2} x^{2} + a^{2} + \frac{a ^{2}}{2} ln x + x^{2} + a^{2} + C

The integrand is real and defined for every real

x

when

a > 0

For $K = \int a^{2} - x^{2} d x$ , differentiation of the root introduces a negative sign. Integration by parts gives $K = x a^{2} - x^{2} + \int x^{2} / a^{2} - x^{2} d x$ . Now use $x^{2} = a^{2} - (a^{2} - x^{2})$ , so $2 K = x a^{2} - x^{2} + a^{2} \int d x / a^{2} - x^{2}$ . The known reciprocal-root integral supplies inverse sine.

Root of a constant minus a squareLaTeX

\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} (\frac{x}{a}) + C

The derivation uses

- a < x < a

. The integrand also has finite endpoint values, unlike its reciprocal.

An alternative route

The substitutions $x = a sec θ$ , $x = a tan θ$ , and $x = a sin θ$ respectively also transform these three radical integrals into trigonometric integrals. Integration by parts shows clearly how each result depends on a reciprocal-root formula already learned.

Shifted radicals use the same three forms

A linear term under the root can be removed by completing the square. After shifting the variable, choose the standard radical formula that matches the sign. The difficult-looking expression is then an application of a derived result rather than a new formula to memorise.

Example — A shifted sum under the root

Problem
Evaluate $\int x^{2} + 2 x + 5 d x$ .

1.Complete the square: $x^{2} + 2 x + 5 = (x + 1)^{2} + 4$ . Put $y = x + 1$ , so $d y = d x$ .
2.Use the sum-of-squares radical formula with $a = 2$ : $\frac{1}{2} y y^{2} + 4 + 2 ln ∣ y + y^{2} + 4 ∣ + C$ .
3.Substitute back to get $\frac{1}{2} (x + 1) x^{2} + 2 x + 5 + 2 ln ∣ x + 1 + x^{2} + 2 x + 5 ∣ + C$ .

Example — A shifted constant-minus-square root

Problem
Evaluate $\int 3 - 2 x - x^{2} d x$ .

1.Write $3 - 2 x - x^{2} = 4 - (x + 1)^{2}$ and put $y = x + 1$ .
2.The constant-minus-square formula gives $\frac{1}{2} y 4 - y^{2} + 2 sin^{- 1} (y /2) + C$ .
3.Thus the answer is $\frac{1}{2} (x + 1) 3 - 2 x - x^{2} + 2 sin^{- 1} ((x + 1) /2) + C$ , on the interior interval $- 3 < x < 1$ .

Common mistake

Integration by parts is a correct identity, but an arbitrary choice of factors need not make an integral solvable by elementary means. Keep the minus sign before the remainder integral. For the radical formulas, the sign of the logarithmic term changes between $x^{2} - a^{2}$ and $x^{2} + a^{2}$ ; derive or differentiate the answer if unsure.

Quick check

Which is the integration-by-parts formula?

Quick check

What is $\int ln x d x$ for $x > 0$ ?

Quick check

Which choice usually simplifies $\int x^{2} e^{x} d x$ ?

Quick check

What is $\int e^{x} [f (x) + f' (x)] d x$ ?

Quick check

Which standard radical integral contains $+ (a^{2} /2) sin^{- 1} (x / a)$ ?

Practice Problems

Find $\int x sin 3 x d x$ .
Find $\int x^{2} ln x d x$ for $x > 0$ .
Find $\int x tan^{- 1} x d x$ .
Find $\int x (ln x)^{2} d x$ for $x > 0$ ; use integration by parts twice.
Find $\int e^{2 x} sin x d x$ by forming and solving an equation for the original integral.
Find $\int e^{x} (1/ x - 1/ x^{2}) d x$ on an interval avoiding zero.
Find $\int 4 - x^{2} d x$ .
Find $\int x^{2} + 4 x + 6 d x$ .
Find $\int x^{2} + 4 x - 5 d x$ on an appropriate interval.
Explain why the coefficient of the logarithm has different signs in the two radical formulas involving $x^{2} + a^{2}$ and $x^{2} - a^{2}$ .

Key Takeaways

Integration by parts reverses the product rule: $\int u d v = uv - \int v d u$ . Choose factors by examining the new integral, and repeat the method when it simplifies the problem. If the original integral returns, collect it algebraically. The exponential product pattern is a direct product derivative. The three radical formulas follow from integration by parts and reciprocal-root formulas; complete the square before applying them to shifted quadratics.

Previous · Lesson 5

Integration by Partial Fractions

Next · Lesson 7

Definite Integrals and the Fundamental Theorem

Reading the product rule backwards

Logarithms, exponentials, and inverse functions

When the original integral returns

Recognising a complete exponential product derivative

Deriving integrals of quadratic square roots

Shifted radicals use the same three forms

Quiz

Practice Problems

Key Takeaways