KindiakMath

The Massive Difference of Plus One

In differential calculus, f(x) and f(x) + 1 yield the same derivative. In integral calculus, a +1 can make all the difference.

Problem 1. Evaluate \displaystyle \int \frac{1}{x^3}\,\mathrm dx.

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Solution. By the reverse power rule,

\displaystyle \int \frac{1}{x^3}\,\mathrm dx = \int x^{-3}\,\mathrm dx = \frac{x^{-2}}{-2} + C = -\frac 1{2x^2} + C.

Problem 2. Evaluate \displaystyle \int \frac{1}{x^3+1}\,\mathrm dx.

(Click for Solution)

Solution. We first note that (-1)^3 + 1 = 0, so that by the factor theorem, x+1 is a factor of x^3+1.

Hence, we factorise x^3+1 = (x+1)(x^2 - x + 1). This suggests us to employ the partial fraction decomposition

\displaystyle \frac 1{x^3 + 1} = A \cdot \frac{1}{x+1} + B \cdot \frac{2x-1}{x^2 - x + 1} + C \cdot \frac{1}{(x-1/2)^2+ (\sqrt{3}/2)^2},

where we will obtain A,B,C later and completed the square in the last term. We recall the following well-known integrals:

\displaystyle \begin{aligned} \int \frac{1}{x+1}\, \mathrm dx &= \ln|x+1|, \\ \int \frac{2x-1}{x^2 - x + 1}\, \mathrm dx &= \ln(x^2-x+1),\\ \int \frac{1}{(x-1/2)^2+ (\sqrt{3}/2)^2}\, \mathrm dx &= \frac{2}{\sqrt 3} \tan^{-1}\left( \frac{2x-1}{\sqrt{3}} \right).\end{aligned}

Integrating on both sides therefore

\displaystyle \int \frac{1}{x^3+1}\, \mathrm dx = A \ln|x+1| + B \ln(x^2-x+1) + C \cdot \frac{2}{\sqrt 3} \tan^{-1}\left( \frac{2x-1}{\sqrt{3}} \right).

All that remains is for us to obtain A, B, C. To that end, clear denominators by multiplying x^3+1 on all sides:

\displaystyle \begin{aligned} 1 &= (A + 2B)x^2 + (-A + B + C)x + (A - B + C). \end{aligned}

Comparing coefficients of x, A = B+C.
Comparing constant terms, 2C =A-B+C = 1, which yields C = 1/2.
Comparing coefficients of x^2, A+2B = 0, which yields A = -2B.

Then, A = B + C implies -2B = B + 1/2 \Rightarrow B = -1/6. Finally A = -2 \cdot -1/6 = 1/3. Substituting and simplifying,

\displaystyle \int \frac{1}{x^3+1}\, \mathrm dx = \frac 13 \ln|x+1| - \frac 16 \ln(x^2-x+1) + \frac{1}{\sqrt 3} \tan^{-1}\left( \frac{2x-1}{\sqrt{3}} \right) + K,

where K is an arbitrary constant.

—Joel Kindiak, 28 Oct 24, 2214H

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  1. Kindiak Gauntlet (Differential Equations) – KindiakMath

    […] Solution. We first note that , so that by the factor theorem, is a factor of . Hence, we factorise . This suggests us to employ the partial fraction decomposition […]

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