Substitution, parts and standard integrals

Choose and combine integration methods, transform bounds and branches consistently, and verify complete antiderivatives by differentiation.

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Look for a derivative you can reverse

An antiderivative must differentiate back to the whole integrand. Before choosing a method, look for a familiar derivative, an inner function with its derivative, a product that becomes simpler, or an identity that changes the expression’s structure. A visible product alone does not decide the method.

For example, xe^x² contains the derivative of the exponent apart from a constant factor. Reversing the chain rule gives

∫ xe^x² dx = (1/2)e^x² + C.

Differentiation checks the factor: (1/2)e^x²(2x) = xe^x². By contrast, the exponent in xe^x has derivative one; its extra factor x is not supplied by that inner derivative. We will use a different reversal for that product.

More generally, on an interval where the real expressions and derivatives exist,

∫ f'(x)[f(x)]ⁿ dx = [f(x)]ⁿ⁺¹/(n + 1) + C, n ≠ -1,

and

∫ f'(x)e^(f(x)) dx = e^(f(x)) + C.

When n = -1, the power formula would divide by zero. Instead,

∫f'(x)/(f(x)) dx = ln |f(x)| + C,

on an interval with f(x) ≠ 0. The absolute value allows a negative inner function while keeping the logarithm real. Separate intervals divided by a zero need not share the same constant.

Thus ∫2x/(x² + 3) dx = ln(x² + 3) + C on the real line, since x² + 3 > 0. Also ∫ dx/(2x-1) = (1/2) ln |2x-1| + C on either side of x = 1/2. Neither formula permits an ordinary definite-integral calculation through a pole.

Check your understanding 1 Match the outer function as well as the inside

Find ∫ x/(x² + 4)² dx and ∫ x/(x² + 4) dx. Explain why one is a reciprocal expression and the other is logarithmic, then differentiate each answer to check its constant factor.

Your two antiderivatives, method distinction and derivative checks, or work on paper

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The common inside is x² + 4, with derivative 2x. The first integrand has outer power -2, giving -1/[2(x² + 4)] + C. Its derivative is (-1/2)(-1)(x² + 4)⁻²(2x) = x/(x² + 4)².

The second has outer power -1, giving (1/2) ln(x² + 4) + C, whose derivative is (1/2)(2x)/(x² + 4). Both integrands exist for every real x. The different denominator powers require different outer antiderivatives.

Only a constant missing factor can be compensated for by multiplying the answer by a constant. Dividing a guessed antiderivative by a variable inner derivative generally adds a new product or quotient contribution when checked. If the entire integrand does not match, rewrite it or change method.

A substitution changes the differential and the bounds

A given substitution replaces the old variable throughout the integral, including its differential. For ∫ dx/square root of (4-x²), use the given substitution x = 2 sin θ with -π/2 < θ < π/2. Then

dx = 2 cos θ dθ, square root of (4-x²) = 2 square root of (cos² θ) = 2 cos θ.

The interval matters: cosine is positive there, so its absolute value can be replaced by cosine. The integral becomes ∫ dθ = θ + C, hence

∫dx/(square root of (4-x²)) = sin⁻¹ (x/2) + C, -2 < x < 2.

Here sin⁻¹ means the inverse sine function, not 1/sin. The final interval is also the integrand’s real domain. Differentiating the inverse sine, including its inner factor 1/2, recovers 1/square root of (4-x²).

For a definite integral, transform both bounds into the new variable, or return completely to the original variable before evaluating at the original bounds. A formula in the new variable with old-variable bounds describes a different calculation.

Check your understanding 2 Keep the integral in one variable

Evaluate I = ∫₀³ x/square root of (x + 1) dx using the given substitution u = square root of (x + 1). Show the transformed integrand and bounds. A calculation uses the correct new integrand but still evaluates between 0 and 3: explain what it missed.

Your substitution, new bounds, exact integral and error explanation, or work on paper

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Since u = square root of (x + 1), we have x = u²-1 and dx = 2u du. Throughout the original interval u > 0, so

x/(square root of (x + 1)) dx = ((u²-1)/u)(2u du) = 2(u²-1) du.

When x = 0, u = 1; when x = 3, u = 2. Thus I = [(2/3)u³-2u]₁² = 8/3. Keeping bounds 0 and 3 would confuse x-values with u-values. An alternative is to substitute back to (2/3)(x + 1)^(3/2)-2 square root of (x + 1) and use x-bounds 0 and 3.

Standard forms depend on scale and denominator signs

For a positive constant a, two inverse-trigonometric forms are

∫dx/(a² + x²) = 1/a tan⁻¹ (x/a) + C,

∫dx/(square root of (a²-x²)) = sin⁻¹ (x/a) + C, |x| < a.

The first exists for all real x; tan⁻¹ means inverse tangent. The second has both a square root and its denominator to respect. The factors and domains follow from differentiation, rather than matching the appearance of a denominator alone.

Completing the square can expose a standard form. Since x² + 4x + 8 = (x + 2)² + 4,

∫dx/(x² + 4x + 8) = 1/2 tan⁻¹ ((x + 2)/2) + C.

Differentiating gives (1/2)[1/(1 + ((x + 2)/2)²)](1/2) = 1/[(x + 2)² + 4]. Forgetting either factor of 1/2 would change the integrand.

The remaining two standard forms are logarithmic:

∫dx/(a²-x²) = 1/(2a) ln |(a + x)/(a-x)| + C,

∫dx/(x²-a²) = 1/(2a) ln |(x-a)/(x + a)| + C.

Use each on an interval excluding x = ± a. For the first, partial fractions give

1/(a²-x²) = (1/(2a))(1/(a + x) + 1/(a-x)).

Integrating the second fraction introduces a minus sign because d(a-x)/dx = -1. Reversing the order of the original denominator negates the integrand, so it must also negate an antiderivative, up to a constant.

Check your understanding 3 Retain the interval and the sign

Find antiderivatives of 1/(4-x²) and 1/(x²-4), and state intervals on which they are valid. Check the sign of the first integrand at zero. Why may neither expression be used by ordinary endpoint substitution for an integral from 0 to 3?

Your two forms, valid intervals and pole check, or work on paper

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The first antiderivative is (1/4) ln |(2 + x)/(2-x)| + C; the second is (1/4) ln |(x-2)/(x + 2)| + C. Each is valid separately on (-∞,-2), (-2,2) or (2,∞), with a constant chosen on that interval. Their derivatives have opposite signs; the first is 1/4 > 0 at zero.

The interval [0,3] crosses the pole at x = 2. Finite values of a logarithmic expression at the two endpoints do not make the intervening integrand continuous or justify an ordinary definite integral through that pole.

Integration by parts reverses a product rule

The product rule says (uv)' = u'v + uv'. Integrating and rearranging gives

∫ u v' dx = uv-∫ u'v dx.

Choose u so that differentiating it helps, and choose v' so that it can actually be integrated. The aim is a simpler remaining integral; swapping the labels without checking that aim may make the work harder.

For ∫ xe^2x dx, take u = x and v' = e^2x. Then u' = 1 and v = (1/2)e^2x, so

∫ xe^2x dx = (x/2)e^2x-(1/4)e^2x + C.

Differentiating gives (1/2)e^2x + xe^2x-(1/2)e^2x = xe^2x. The two extra terms cancel. Trying to integrate the two factors independently and multiply their answers would not have this cancellation.

For ∫ ln x dx on x > 0, regard the integrand as (ln x) · 1. Choose u = ln x, v' = 1, so u' = 1/x, v = x. Then

∫ ln x dx = x ln x-∫1 dx = x ln x-x + C.

The logarithm becomes simpler when differentiated even though it was not initially written as a product. Its positive-input restriction stays with the answer.

Check your understanding 4 Choose a product that becomes simpler

Find ∫ x cos x dx. Show your choice of u and v', and differentiate the complete answer to check its sign.

Your method choice, antiderivative and derivative check, or work on paper

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Choose u = x, v' = cos x, so u' = 1, v = sin x. The integral is x sin x-∫ sin x dx = x sin x + cos x + C. Differentiation gives sin x + x cos x- sin x = x cos x. The cosine term is positive because the subtracted integral of sine is -(- cos x).

Sometimes one application is not enough. For ∫ x²e^x dx, taking u = x² gives x²e^x-2∫ xe^x dx. A second application gives e^x(x²-2x + 2) + C. Its derivative is e^x[(x²-2x + 2) + (2x-2)] = x²e^x. Each application lowers the polynomial degree.

Rewrite trigonometric powers before integrating

All angles here are in radians. A square of a trigonometric function is not generally integrated by squaring its antiderivative. Double-angle identities instead give

∫ sin² x dx = ∫(1- cos 2x)/2 dx = x/2-(sin 2x)/4 + C,

∫ cos² x dx = x/2 + (sin 2x)/4 + C.

The derivative of sin 2x supplies a factor of two. For tangent, use tan² x = sec² x-1 to obtain ∫ tan² x dx = tan x-x + C on an interval with cos x ≠ 0.

Check your understanding 5 Use the identity inside the bounds

Evaluate ∫₀^(π/4) tan² x dx exactly. Explain why its sign is consistent with the original integrand and why ∫ tan² x dx is not (∫ tan x dx)².

Your exact integral and derivative-based explanation, or work on paper

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Using tan² x = sec² x-1 gives [tan x-x]₀^(π/4) = 1-π/4 > 0. Tangent squared is nonnegative and positive inside this interval. Squaring an antiderivative introduces a product-rule factor when differentiated, so it does not generally recover the square of the integrand.

Problem

Use the given substitution u = x² to evaluate J = ∫₀¹ x³e^x² dx exactly. Choose how to finish after changing variable, and verify an antiderivative in the original variable. A proposed answer uses (1/2)e^x² as the antiderivative because the exponent has derivative 2x. Explain what the proposal overlooks.

Write your answer, then compare it with the solution

Your transformed integral, method, exact value and check, or work on paper

Hints

Hint 1

Account for every factor after changing variable

Write x³ dx = x²(x dx) and transform the two endpoints. Decide whether the remaining integral is a direct inner-derivative match or needs another method.

Show full solution

Transform the complete integrand
Method
Replace the polynomial factors, exponential, differential and bounds together.
Reason
The given substitution absorbs x dx, leaving another factor that still matters.
Working
Answer
Since u = x² and du = 2x dx, x³ dx = (u/2) du. The x-bounds 0,1 become u-bounds 0,1, so J = (1/2)∫₀¹ue^u du. These bounds happen to agree numerically; they have still been transformed.
Choose a simpler product and evaluate
Method
Integrate ue^u by parts, differentiating the polynomial factor.
Reason
The factor u does not come from differentiating the exponent u, and it becomes simpler when differentiated.
Working
Answer
∫ ue^u du = ue^u-e^u + C = (u-1)e^u + C. Therefore J = (1/2)[(u-1)e^u]₀¹ = (1/2)[0-(-1)] = 1/2.
Check in the original variable
Method
Differentiate F(x) = (1/2)(x²-1)e^x².
Reason
A correct transformed calculation must recover the entire original integrand.
Working
Answer
F'(x) = xe^x² + x(x²-1)e^x² = x³e^x². The proposed (1/2)e^x² differentiates only to xe^x², missing a factor x². Also F(1)-F(0) = 0-(-1/2) = 1/2, agreeing with the transformed evaluation.

Quick check

The substitution and the finishing method do different jobs. Compare every transformed factor and endpoint, then check the complete original integrand by differentiation. Equivalent antiderivatives may look different; on the same interval they can differ by a constant.

Choose a method from the complete expression, keep its real-domain and branch conditions, and finish in the requested variable. An indefinite integral needs a constant; a definite integral needs consistent bounds. Differentiation is the final check on factors, signs and missing terms.

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Substitution, parts and standard integrals

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Reading settings

Look for a derivative you can reverse

Check your understanding 1 Match the outer function as well as the inside

A substitution changes the differential and the bounds

Check your understanding 2 Keep the integral in one variable

Standard forms depend on scale and denominator signs

Check your understanding 3 Retain the interval and the sign

Integration by parts reverses a product rule

Check your understanding 4 Choose a product that becomes simpler

Rewrite trigonometric powers before integrating

Check your understanding 5 Use the identity inside the bounds

Evaluate and verify the whole integral

Transform the complete integrand

Choose a simpler product and evaluate

Check in the original variable

Keep track of this lesson