Porisma

Worked solution

Use logarithms after matching or division

Integrate with respect to xx: (a) 12x+7\frac1{2x+7}; (b) x2+2x3+6x\frac{x^2+2}{x^3+6x}; (c) x2x+1\frac{x^2}{x+1}. How do we handle integrals containing improper rational functions?

CJC · Lecture · Example 3

(a)

With f(x)=2x+7f(x)=2x+7 and f′(x)=2f^{\prime}(x)=2, ∫12x+7 dx=12∫22x+7 dx=12ln⁡∣2x+7∣+C\int\frac1{2x+7}\,dx=\frac12\int\frac2{2x+7}\,dx=\frac12\ln|2x+7|+C, for x≠−7/2x\ne-7/2.

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(b)

With f(x)=x3+6xf(x)=x^3+6x and f′(x)=3x2+6=3(x2+2)f^{\prime}(x)=3x^2+6=3(x^2+2), ∫x2+2x3+6x dx=13∫3(x2+2)x3+6x dx=13ln⁡∣x3+6x∣+C\int\frac{x^2+2}{x^3+6x}\,dx=\frac13\int\frac{3(x^2+2)}{x^3+6x}\,dx=\frac13\ln|x^3+6x|+C, for x≠0x\ne0.

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(c) Divide first

The Teacher answer to the improper-rational-function question is long division. Divide x2x^2 by x+1x+1: subtract x(x+1)x(x+1) to leave −x-x; subtract −(x+1)-(x+1) to leave remainder 11. Hence x2x+1=x−1+1x+1\frac{x^2}{x+1}=x-1+\frac1{x+1} and ∫x2x+1 dx=12x2−x+ln⁡∣x+1∣+C\int\frac{x^2}{x+1}\,dx=\frac12x^2-x+\ln|x+1|+C, for x≠−1x\ne-1.

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Common mistakes

Retain the modulus in logarithms unless the argument is known positive. Work on intervals that do not cross a zero of the original denominator.

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