Question

17. Using method of partial fraction to evaluate S (x + 5) dx/(x + 1) (x + 2)2(a) 4 log (x + 1) – 4 log (x + 2) + 3/x + 2+ k(b) 4 log (x + 2) - 3/x + 2+k(c) 4 log (x + 1) - 4 log (x + 2)(d) none of these

Answer 1

Step-by-step explanation:

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Section 1-4 : Partial Fractions

In this section we are going to take a look at integrals of rational expressions of polynomials and once again let’s start this section out with an integral that we can already do so we can contrast it with the integrals that we’ll be doing in this section.

∫2x−1x2−x−6dx=∫1uduusing u=x2−x−6 and du=(2x−1)dx=ln∣∣x2−x−6∣∣+c∫2x−1x2−x−6dx=∫1uduusing u=x2−x−6 and du=(2x−1)dx=ln⁡|x2−x−6|+c

So, if the numerator is the derivative of the denominator (or a constant multiple of the derivative of the denominator) doing this kind of integral is fairly simple. However, often the numerator isn’t the derivative of the denominator (or a constant multiple). For example, consider the following integral.

∫3x+11x2−x−6dx∫3x+11x2−x−6dx

In this case the numerator is definitely not the derivative of the denominator nor is it a constant multiple of the derivative of the denominator. Therefore, the simple substitution that we used above won’t work. However, if we notice that the integrand can be broken up as follows,

3x+11x2−x−6=4x−3−1x+23x+11x2−x−6=4x−3−1x+2

then the integral is actually quite simple.

∫3x+11x2−x−6dx=∫4x−3−1x+2dx=4ln|x−3|−ln|x+2|+c∫3x+11x2−x−6dx=∫4x−3−1x+2dx=4ln⁡|x−3|−ln⁡|x+2|+c

This process of taking a rational expression and decomposing it into simpler rational expressions that we can add or subtract to get the original rational expression is called partial fraction decomposition. Many integrals involving rational expressions can be done if we first do partial fractions on the integrand.

So, let’s do a quick review of partial fractions. We’ll start with a rational expression in the form,

f(x)=P(x)Q(x)f(x)=P(x)Q(x)

where both P(x)P(x) and Q(x)Q(x) are polynomials and the degree of P(x)P(x) is smaller than the degree of Q(