Why These Mixed Integrals Are Manageable
A product of a linear polynomial and a logarithm or inverse-trigonometric function may look difficult because neither factor is itself a simple power of \(x\). The useful feature is that differentiating the logarithmic or inverse-trigonometric factor usually produces a simpler expression, while integrating the linear polynomial produces a quadratic.
Integration by parts puts those two changes together. The method does not guarantee that the remaining integral is immediately simple, so the choice of parts must be followed by careful algebra and a check that the original domain is respected.
Formula and Conditions
For differentiable functions \(u\) and \(v\), the product rule gives \((uv)'=u'v+uv'\). Rearranging and integrating yields the integration-by-parts formula:
Here \(dv\) is a differential expression with an antiderivative \(v\), and \(du=u'(x)\,dx\). For an indefinite integral, the identity applies on an interval where the functions and derivatives involved are defined and the remaining integral exists. The final antiderivative is understood up to an arbitrary constant \(C\).
If the linear factor is \(ax+b\), where \(a\) and \(b\) are constants, its antiderivative is \(\frac{a}{2}x^2+bx\). Thus, for a factor \(g(x)\) with a defined derivative, the first integration-by-parts step is often
The new integral may contain a rational expression or a square-root expression. That is the point at which simplification matters: algebraic rearrangement can reveal terms with familiar antiderivatives.
Worked Example: A Linear Polynomial Times a Logarithm
Worked Example: A Linear Polynomial Times a Logarithm
Evaluate \(\int (2x+1)\ln(x+2)\,dx\), and check the result.
The logarithm requires \(x+2>0\), so work on the interval \(x>-2\). Choose \(u=\ln(x+2)\) and \(dv=(2x+1)\,dx\). Then
Integration by parts gives
Divide the numerator by the denominator. Since \((x+2)(x-1)=x^2+x-2\), the remainder is \(2\), and therefore
Integrating the three terms gives
Consequently, an antiderivative is
To check it, differentiate the two logarithmic terms using the product rule:
The derivative recovers the integrand, and the logarithm is defined throughout the stated interval.
Worked Example: A Linear Polynomial Times an Inverse Tangent
Worked Example: A Linear Polynomial Times an Inverse Tangent
Evaluate \(\int(3x-2)\arctan(2x)\,dx\), and check the result.
The inverse tangent is defined for every real \(x\). Choose \(u=\arctan(2x)\) and \(dv=(3x-2)\,dx\). The chain rule gives
Therefore,
For the remaining integral, write \(3x^2=\frac{3}{4}(1+4x^2)-\frac{3}{4}\). Then
Integrate each term. The inverse-tangent term follows from \(\frac{d}{dx}\arctan(2x)=\frac{2}{1+4x^2}\); the logarithmic term follows by differentiating \(\ln(1+4x^2)\). Thus
Substituting this expression into the integration-by-parts result gives
For the check, set \(v=\frac{3}{2}x^2-2x\). The expression can also be written \(F=v\arctan(2x)-R+C\), where \(R\) is the displayed antiderivative of the remaining rational expression. Its derivative is \(R'=(3x^2-4x)/(1+4x^2)=2v/(1+4x^2)\). Hence
because \(2v=3x^2-4x\). This verifies the antiderivative. Also, \(1+4x^2>0\) for every real \(x\), so the logarithm in the result is defined on the whole real line.
Worked Example: A Linear Polynomial Times an Inverse Sine
Worked Example: A Linear Polynomial Times an Inverse Sine
Evaluate \(\int(x+1)\arcsin(x/2)\,dx\) on its interior domain, and check the result.
The inverse sine requires \(-1\le x/2\le1\), or \(-2\le x\le2\). Its derivative is finite for \(-2<x<2\), so carry out integration by parts on that open interval. Let \(u=\arcsin(x/2)\) and \(dv=(x+1)\,dx\). Then
Integration by parts gives
To evaluate the remaining integral, split its numerator. Since \(x^2=4-(4-x^2)\),
The first term has antiderivative \(2\arcsin(x/2)\). For the second, use \(\int\sqrt{4-x^2}\,dx=\frac{x}{2}\sqrt{4-x^2}+2\arcsin(x/2)\). For the third, the substitution \(w=4-x^2\), \(dw=-2x\,dx\), gives antiderivative \(-\sqrt{4-x^2}\). Combining these pieces,
Subtracting this from the product term yields
To verify the result, let \(S=\sqrt{4-x^2}\), \(A=\frac{x^2}{2}+x-1\), and \(B=\frac{x}{4}+1\). Then \(A'=x+1\), \(\frac{d}{dx}\arcsin(x/2)=1/S\), and \(S'=-x/S\). Differentiating gives
Since \(A-Bx=-S^2/4\), the last two terms cancel: \(\frac{A-Bx}{S}+\frac{S}{4}=-S/4+S/4=0\). Thus \(F'(x)=(x+1)\arcsin(x/2)\), as required.
Why the Choice of Parts Works
The method is justified by the product rule, not by a guess about which factor looks more complicated. The choice is useful because it changes the factors in complementary ways: the logarithmic or inverse-trigonometric factor is differentiated, and the linear polynomial is integrated. For example, \(\ln(x+2)\) becomes \(1/(x+2)\), while a linear polynomial becomes a quadratic. The resulting rational expression can often be simplified by division or by separating terms.
For an inverse-trigonometric factor, its derivative introduces a denominator such as \(1+x^2\), \(1+4x^2\), or \(\sqrt{4-x^2}\). The quadratic produced by integrating the linear factor can often be rearranged to match that denominator. In the inverse-sine example, rewriting \(x^2\) as \(4-(4-x^2)\) exposed both an inverse-sine antiderivative and a square-root antiderivative.
Common Mistake: Losing a Chain-Rule Factor
When differentiating \(\arctan(2x)\), it is incorrect to write \(du=dx/(1+4x^2)\). The chain rule contributes a factor of \(2\): the correct differential is \(du=2\,dx/(1+4x^2)\). The error changes the remaining integral in the second example.
In that example, let \(v=\frac{3}{2}x^2-2x\), so \(v'=3x-2\). If one incorrectly uses \(du=dx/(1+4x^2)\), the proposed integration-by-parts expression is \(G=v\arctan(2x)-\int v/(1+4x^2)\,dx\). Differentiating it shows exactly why it fails:
The extra term is generally nonzero, so \(G'\) is not the original integrand. The correction is to differentiate the composite function \(\arctan(2x)\) with the chain rule before applying integration by parts, then check the completed answer by differentiation.
Check Your Understanding
Use integration by parts, keep track of the domain, and differentiate your proposed answer to check it.
- For \(\int (x+2)\ln(x+1)\,dx\), what interval makes the logarithm defined, and what are suitable choices of \(u\) and \(dv\)?
- What derivative of \(\arctan(3x)\) must be used when applying integration by parts?
- In \(\int x\arcsin(x/3)\,dx\), what is the interior interval on which the inverse-sine derivative is finite?
- Why can simplifying the remaining rational expression be more useful than trying to integrate it in its original form?
- What check detects a missing chain-rule factor in an integration-by-parts calculation?
