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Antiderivatives and Basic Integrals · Tutorial 1082 of 1495

Mixed Basic Antiderivative Problems

Practise selecting and combining elementary antiderivative rules, then verify each result by differentiating.

Beginner 9 min read

What You'll Learn

  • Apply the power rule while recognizing why the reciprocal case must be handled separately.
  • Find antiderivatives of reciprocal functions and logarithmic functions on their appropriate domains.
  • Account for constant factors inside exponential and trigonometric functions.
  • Combine antiderivatives of sums using linearity.
  • Check a proposed antiderivative by differentiating it term by term.
  • Identify and correct common exponent and coefficient errors.

Introduction

A mixed antiderivative problem asks you to integrate a sum of familiar kinds of functions. The main task is not to find one new rule for the whole expression. It is to recognize the individual terms, select the right rule for each, and combine the results carefully.

This tutorial brings together basic rules for powers, reciprocal functions, logarithms, exponentials, and trigonometric functions. The most reliable check is to differentiate your proposed answer: the derivative should reproduce the original integrand.

Antiderivatives and Indefinite Integrals

A function \(F\) is an antiderivative of \(f\) on an interval if \(F'(x)=f(x)\) at every point of that interval. The notation \(\int f(x)\,dx\) represents the family of antiderivatives of \(f\). On an interval, any two antiderivatives of the same function differ by a constant, so the general answer includes \(+C\).

Definition: If \(F'(x)=f(x)\) throughout an interval, then \(F\) is an antiderivative of \(f\) there. We write \(\int f(x)\,dx=F(x)+C\), where \(C\) is an arbitrary constant.

The interval matters. For example, \(1/x\) is not defined at \(x=0\), so an antiderivative formula for \(1/x\) applies on intervals that do not contain zero. Likewise, \(\ln x\) is defined for positive \(x\).

A Rule for Each Kind of Term

For a real exponent \(n\ne -1\), the power rule for antiderivatives is

$$ \int x^n\,dx=\frac{x^{n+1}}{n+1}+C. $$

The restriction \(n\ne -1\) is essential: when \(n=-1\), the denominator \(n+1\) would be zero. That case is the reciprocal rule:

$$ \int \frac{1}{x}\,dx=\ln|x|+C,\qquad x\ne 0. $$

The absolute value allows the formula to apply on either interval \(x>0\) or \(x<0\), since \(\frac{d}{dx}\ln|x|=1/x\) on both. For a logarithmic integrand on the positive domain, a useful formula is

$$ \int \ln x\,dx=x\ln x-x+C,\qquad x>0. $$

The remaining basic formulas include exponentials and trigonometric functions. A constant multiplying \(x\) inside the function changes the antiderivative by a corresponding factor in the denominator.

$$ \int e^{kx}\,dx=\frac{1}{k}e^{kx}+C\quad(k\ne 0), $$ $$ \int \cos(kx)\,dx=\frac{1}{k}\sin(kx)+C\quad(k\ne 0), $$ $$ \int \sin(kx)\,dx=-\frac{1}{k}\cos(kx)+C\quad(k\ne 0). $$

When \(k=1\), these formulas give the usual antiderivatives of \(e^x\), \(\cos x\), and \(\sin x\). The constant-factor versions can be checked by differentiating the proposed answers.

Why the Rules Work Together

The sum rule for derivatives explains why a sum of separate antiderivatives can be combined. If \(F'=f\) and \(G'=g\), then

$$ (F+G)'=F'+G'=f+g. $$

More generally, for constants \(a_1,\ldots,a_m\), if \(F_i'=f_i\) for each \(i\), then

$$ \left(\sum_{i=1}^{m}a_iF_i\right)' =\sum_{i=1}^{m}a_iF_i' =\sum_{i=1}^{m}a_if_i. $$

This proves the linearity rule for antiderivatives: integrate each term separately, keep its constant multiplier, and add the results. One arbitrary \(+C\) is enough for the combined answer.

The power formula also follows directly from differentiation. For \(n\ne -1\), the power rule for derivatives gives

$$ \frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right) =\frac{1}{n+1}(n+1)x^n =x^n. $$

Thus the proposed expression is an antiderivative of \(x^n\). The same checking idea applies to each basic formula: differentiate the proposed answer and confirm that it returns the integrand.

A Practical Method for Mixed Problems

For a sum of basic terms, use the following sequence. First identify the terms and their domains. Then match each term to a formula, paying particular attention to reciprocal terms and to constants multiplying \(x\) inside a function. Finally, differentiate the whole proposed answer.

1
Separate the terms.
Use linearity to treat each summand individually.
2
Choose a rule for each term.
Distinguish ordinary powers from the reciprocal case, and account for any constant factor inside an exponential or trigonometric function.
3
Combine and verify.
Add the antiderivatives, include \(+C\), and differentiate term by term to check the result.

Worked Examples

Worked Example: Powers, a Reciprocal, and a Logarithm

Find \(\int (4x^3-6/x+2\ln x)\,dx\) for \(x>0\).

Separate the three terms and apply the corresponding formulas:

$$ \int \left(4x^3-\frac{6}{x}+2\ln x\right)\,dx =4\int x^3\,dx-6\int \frac{1}{x}\,dx+2\int\ln x\,dx. $$

For the power term, \(4\int x^3\,dx=4(x^4/4)=x^4\). Since \(x>0\), \(\ln|x|=\ln x\), so the reciprocal term contributes \(-6\ln x\). The logarithm term contributes \(2(x\ln x-x)\). Therefore,

$$ \int \left(4x^3-\frac{6}{x}+2\ln x\right)\,dx =x^4-6\ln x+2x\ln x-2x+C. $$

Check by differentiation:

$$ \frac{d}{dx}\left(x^4-6\ln x+2x\ln x-2x\right) =4x^3-\frac{6}{x}+2(\ln x+1)-2 =4x^3-\frac{6}{x}+2\ln x. $$

The derivative matches the integrand, so the answer is verified.

Worked Example: Exponentials and Trigonometric Functions

Find \(\int (3e^{3x}-4\cos(2x)+5\sin x)\,dx\).

Integrate each term. For the first term, the factor \(3\) outside cancels the \(1/3\) from integrating \(e^{3x}\). For the second, the inside factor \(2\) gives a \(1/2\):

$$ \begin{aligned} \int \left(3e^{3x}-4\cos(2x)+5\sin x\right)\,dx &=3\left(\frac{1}{3}e^{3x}\right) -4\left(\frac{1}{2}\sin(2x)\right) +5(-\cos x)+C\\ &=e^{3x}-2\sin(2x)-5\cos x+C. \end{aligned} $$

Differentiate to check:

$$ \frac{d}{dx}\left(e^{3x}-2\sin(2x)-5\cos x\right) =3e^{3x}-4\cos(2x)+5\sin x. $$

Each coefficient returns to its required value, including the factor \(2\) from differentiating \(\sin(2x)\).

Worked Example: Combining All Five Kinds of Terms

Find an antiderivative of \(2x^2-3/x+4\ln x+5e^{2x}+6\cos x-2\sin x\) for \(x>0\).

Match each term in order: the power rule applies to \(2x^2\); the reciprocal rule to \(-3/x\); the logarithm formula to \(4\ln x\); the exponential formula with \(k=2\) to \(5e^{2x}\); and the sine and cosine formulas to the final two terms.

$$ \begin{aligned} \int \left(2x^2-\frac{3}{x}+4\ln x+5e^{2x}+6\cos x-2\sin x\right)\,dx &=\frac{2}{3}x^3-3\ln x+4(x\ln x-x)\\ &\quad+\frac{5}{2}e^{2x}+6\sin x+2\cos x+C. \end{aligned} $$

The last term has a positive sign because \(\int\sin x\,dx=-\cos x\), and its coefficient is \(-2\). Now differentiate:

$$ \begin{aligned} \frac{d}{dx}\left(\frac{2}{3}x^3-3\ln x+4x\ln x-4x+\frac{5}{2}e^{2x}+6\sin x+2\cos x\right) &=2x^2-\frac{3}{x}+4(\ln x+1)-4\\ &\quad+5e^{2x}+6\cos x-2\sin x\\ &=2x^2-\frac{3}{x}+4\ln x+5e^{2x}+6\cos x-2\sin x. \end{aligned} $$

The derivative is exactly the original integrand. The constant \(C\) is omitted from the differentiation line because its derivative is zero.

Common Pitfall: Applying the Power Rule to \(1/x\)

A tempting but invalid calculation is

$$ \int x^{-1}\,dx \overset{\text{wrong}}{=} \frac{x^0}{0}. $$

The power-rule formula requires \(n\ne -1\), and here \(n+1=0\). The expression on the right is undefined, so the power rule does not apply. Use the reciprocal rule instead:

$$ \int x^{-1}\,dx=\ln|x|+C,\qquad x\ne 0. $$

Another frequent error is to omit a factor from the inside of an exponential or trigonometric function. For example, \(\int e^{3x}\,dx\) is not \(e^{3x}+C\), because differentiating \(e^{3x}\) gives \(3e^{3x}\). The correct antiderivative is \(\frac13e^{3x}+C\).

Why Careful Rule Selection Matters

Mixed expressions are useful practice because they require more than recalling isolated formulas. A correct solution depends on recognizing the exceptional reciprocal case, preserving signs, and accounting for constants inside functions. Keeping the terms separate until each rule has been applied makes those choices visible and makes errors easier to locate.

The final differentiation check is especially valuable. It tests the entire result, not merely one remembered formula. If the derivative has the wrong sign or coefficient, compare it term by term with the original integrand and correct the matching antiderivative.

Check Your Understanding

For each question, identify the rule that applies and check any proposed antiderivative by differentiation.

  1. Find \(\int (5x^4+2/x)\,dx\) on \(x>0\).
  2. Find \(\int (4e^{4x}-3\cos(3x))\,dx\).
  3. Why does the power-rule formula not apply to \(x^{-1}\), and what formula replaces it?
  4. Find an antiderivative of \(\ln x\) for \(x>0\), then verify it.
  5. In a sum of several basic functions, why may the antiderivatives be added term by term?

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