Why the Natural Logarithm Has a Less Obvious Antiderivative
For many basic functions, an antiderivative can be recognized by recalling a familiar differentiation rule. The natural logarithm presents a small puzzle: its derivative is \(1/x\), but the function \(\ln x\) itself is not a simple power of \(x\). The power rule for antiderivatives therefore does not directly give its antiderivative.
A useful approach is to look for a product whose derivative contains \(\ln x\). The expression \(x\ln x\) is a natural candidate: differentiating it uses the product rule, and one of the resulting terms is exactly \(\ln x\). The other term can then be accounted for.
Domain and Definition
The real natural logarithm \(\ln x\) is defined only when \(x>0\). Accordingly, the antiderivative formula in this tutorial applies on the positive real numbers. It does not describe a real-valued antiderivative on an interval containing zero or negative inputs.
Recall that an antiderivative of a function \(f\) on an interval is a differentiable function \(F\) such that \(F'(x)=f(x)\) at every point in that interval. Indefinite integral notation records the family of antiderivatives. In this case, we seek a function \(F\) satisfying \(F'(x)=\ln x\) for every \(x>0\).
The needed differentiation facts are the product rule and the derivative of the natural logarithm:
Reverse the Product Rule
Start with the product \(x\ln x\). The product rule gives two terms. One is the desired logarithm, and the other is a constant:
Thus \(x\ln x\) differentiates to \(\ln x+1\), not just \(\ln x\). To remove the extra \(1\), subtract a function whose derivative is \(1\), namely \(x\). This suggests \(x\ln x-x\). The computation is not a guess that must be trusted: differentiation checks it exactly.
This is the product rule used in reverse. In forward differentiation, a product produces a sum of terms. Here, we identify a product that produces the desired function along with an extra term, then cancel that extra term by subtracting its antiderivative.
Derivation and Verification of the Formula
The derivation can be written as a complete chain of reasoning. First, the product rule and \((\ln x)'=1/x\) establish that the derivative of \(x\ln x\) is \(\ln x+1\) for \(x>0\). Second, the derivative of \(x\) is \(1\). Subtracting these derivative results yields the derivative of the difference:
Every equality holds for \(x>0\), where \(\ln x\) is defined. Therefore \(x\ln x-x\) is an antiderivative of \(\ln x\) on \((0,\infty)\). The earlier result that any two antiderivatives of the same function on an interval differ by a constant then gives the full family: every antiderivative on this interval has the form \(x\ln x-x+C\).
The constant \(C\) appears because differentiating a constant gives zero. Adding any constant to a verified antiderivative does not change its derivative. Conversely, the difference-between-antiderivatives result ensures that this accounts for all antiderivatives on the interval, rather than only providing one example.
Worked Examples
Worked Example: Find an Antiderivative of \(\ln x\)
Find \(\int \ln x\,dx\) for \(x>0\), and verify the result.
The product-rule calculation gives the candidate \(x\ln x-x\). Include the arbitrary constant to represent the full family:
Now differentiate the proposed antiderivative. Applying the product rule to \(x\ln x\) gives
The derivative equals the integrand for every \(x>0\), so the formula is verified. The constant contributes zero to the derivative.
Worked Example: Integrate a Constant Multiple and a Constant
Find an antiderivative of \(4\ln x+3\) on \((0,\infty)\).
By linearity, integrate the two terms separately. The antiderivative of \(\ln x\) is \(x\ln x-x\), and an antiderivative of the constant \(3\) is \(3x\):
Check by differentiating the simplified expression:
The derivative matches the original integrand, so \(4x\ln x-x+C\) is the required family of antiderivatives for \(x>0\).
Worked Example: Check a Proposed Antiderivative
Determine whether \(G(x)=2x\ln x-2x+7\) is an antiderivative of \(2\ln x\) on \((0,\infty)\).
Differentiate each term. For the product, use the product rule:
Thus the derivative is the stated function for every positive \(x\). The constant \(7\) does not affect the derivative, so \(G\) is indeed an antiderivative of \(2\ln x\) on \((0,\infty)\).
As a numerical check at \(x=1\), \(\ln 1=0\), so \(G'(1)=2\ln 1=0\). The original integrand at \(x=1\) is also \(2\ln 1=0\), consistent with the general calculation.
A Common Mistake: Forgetting the Extra Term
A frequent error is to notice that \(x\ln x\) contains \(\ln x\) and conclude that it must be an antiderivative of \(\ln x\). That overlooks the second term generated by the product rule. The incorrect calculation would be:
The correct calculation includes both product-rule terms:
Therefore, \(x\ln x\) alone is not an antiderivative of \(\ln x\). Subtracting \(x\) corrects the extra \(1\), because
A second error is to write this real-valued formula for \(x\leq 0\). The expression \(\ln x\) is not defined as a real number there, so the real antiderivative statement must retain the condition \(x>0\). The absolute value used in the earlier antiderivative of the reciprocal function addresses a different integrand; it does not make \(\ln x\) defined for negative \(x\).
How to Recognize the Method
When an integrand is \(\ln x\), the useful clue is that differentiating \(x\ln x\) creates \(\ln x\) as one of its terms. A reliable reasoning sequence is:
Use \(x\ln x\), since its product rule includes the desired factor \(\ln x\).
The result is \(\ln x+1\), not merely \(\ln x\).
Subtract \(x\), whose derivative is \(1\), to obtain \(x\ln x-x\).
Differentiate the candidate to recover \(\ln x\), and specify \(x>0\).
The point of the method is not to memorize a formula without checking it. The product rule explains why the term \(x\ln x\) appears, and differentiation confirms why the correction \(-x\) is required. This makes the formula easier to reconstruct and less likely to be confused with the antiderivative of a different logarithmic expression.
Summary
On positive inputs, the product rule gives \(\frac{d}{dx}(x\ln x)=\ln x+1\). Removing the extra \(1\) by subtracting \(x\) yields an antiderivative of the natural logarithm. The full family on \((0,\infty)\) is
The essential checks are to include both terms from the product rule, retain the domain condition \(x>0\), and differentiate the proposed answer to confirm that it returns the integrand.
Check Your Understanding
Answer these questions by using the product rule and checking proposed antiderivatives by differentiation.
- What is the derivative of \(x\ln x\) for \(x>0\), and which term must be corrected to obtain \(\ln x\)?
- Why does subtracting \(x\) from \(x\ln x\) produce an antiderivative of \(\ln x\)?
- Find the family of antiderivatives of \(3\ln x-2\) on \((0,\infty)\), and verify your result by differentiation.
- Is \(x\ln x+C\) an antiderivative of \(\ln x\)? Explain by computing its derivative.
- Why must the real-valued formula be stated for \(x>0\)?
