Integration by parts and partial fractions

Integration by parts and partial fractions solve different kinds of problems. Integration by parts is useful when an integrand contains a product whose structure becomes simpler after one factor is differentiated, and partial fractions is useful for rational functions that can be decomposed into simpler fractions. The first step is recognizing which structure is present.

Integration by parts

The product rule says \(\tfrac{d}{dx}(uv) = u\tfrac{dv}{dx} + v\tfrac{du}{dx}\), so integrating both sides gives \(\textstyle uv = \int u\,dv + \int v\,du\). Rearranging,

\[\int u\,dv = uv-\int v\,du\]

This is integration by parts. The choice of \(u\) matters, and a useful choice is one whose derivative is simpler than the original factor.

Consider \(\textstyle\int xe^x\,dx\). Choose \(u=x,\; dv=e^x\,dx\), then \(du=dx,\; v=e^x\), so

\[\int xe^x\,dx = xe^x-\int e^x\,dx = xe^x-e^x+C\]

The remaining integral is simpler than the one we started with. If instead we choose \(u=e^x,\; dv=x\,dx\), the resulting integral becomes more complicated. The formula is valid either way, and the useful choice is the one that improves the problem.

A function that looks like one factor

Integration by parts can also be used when there is no visible product. For \(\textstyle\int \ln x\,dx\), write \(\ln x = (\ln x)(1)\), choose \(u=\ln x,\; dv=dx\), and then \(du=\tfrac1x\,dx,\; v=x\). Therefore \(\textstyle\int \ln x\,dx = x\ln x-\int1\,dx = x\ln x-x+C\), and the invisible factor of 1 is what makes the product-rule structure available.

Partial fractions

Partial fractions begins with a rational function. Suppose \(\tfrac{1}{x^2-1} = \tfrac{1}{(x-1)(x+1)}\), and we seek constants \(A\) and \(B\) such that

\[\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}\]

Multiplying through by the denominator gives \(1 = A(x+1)+B(x-1)\), and setting \(x=1,\; x=-1\) in turn gives \(A=\tfrac12,\; B=-\tfrac12\). So \(\tfrac{1}{x^2-1} = \tfrac12\tfrac{1}{x-1} - \tfrac12\tfrac{1}{x+1}\). Now integrate term by term:

\[\int\frac{1}{x^2-1}\,dx = \frac12\ln\vert x-1\vert - \frac12\ln\vert x+1\vert + C\]

The original rational function did not have an obvious antiderivative. After decomposition, each term does. For the course-level problems considered here, the denominator factors into distinct linear factors, and more general decompositions can involve repeated factors and irreducible quadratics, but those require additional forms.

Check by differentiating

An antiderivative can be checked immediately. Differentiate \(\tfrac12\ln\vert x-1\vert - \tfrac12\ln\vert x+1\vert\) and the result is \(\tfrac{1}{2(x-1)} - \tfrac{1}{2(x+1)} = \tfrac{1}{x^2-1}\).

This is a useful habit after any longer integration problem. It is especially important with definite integrals because a wrong antiderivative can occasionally produce a plausible numerical answer for particular bounds, and a correct final value does not repair incorrect reasoning.

Choosing the technique

A useful classification is:

• If the integrand contains a composite function and its derivative, try substitution.
• If it contains a product where differentiating one factor makes the expression simpler, consider integration by parts.
• If it is a rational function with a factorable denominator, consider partial fractions.