• Calculus > Partial Fractions

Decomposing Rational Functions into Partial Fractions

INTRODUCTION

Partial fractions can help us integrate rational functions much more easily that tackling integration of the rational function as is. Rational functions of the form \( f(x) = \frac{P(x)}{Q(x)} \) can be decomposed into partial fractions with a simple numerator and denominator.

Most textbooks show the manual calculation of finding the numerators of partial fractions using system of equations. For example, lamar.edu site has several examples with manual calculations. However, this page presents an easier, sure-fire method.

We let \( Q(x) \) be a polynomial of degree n and \( P(x) \) be a polynomial of degree less than n.

Next, let’s assume \( Q(x) \) factors as \( Q(x) = (x - r_{1})(x - r_{2})(x - r_{3})...(x - r_{n}) \), where \( r_{n} \) are the roots of \( Q(x) \). Then, we will decompose the rational function as follows:

(i) \( \frac{P(x)}{Q(x)} = \frac{K_{1}}{x - r_{1}} + \frac{K_{2}}{x - r_{2}} + \frac{K_{3}}{x - r_{3}} + ... + \frac{K_{n}}{x - r_{n}} \) for some constants \( K_{i} \) in the numerators.

The number of terms will depend on the degree of \( Q(x) \).

The denominator in RHS of (i) is simply \( Q(x) = (x - r_{1})(x - r_{2})(x - r_{3})...(x - r_{n}) \) and the numerator is as follows if we multiplied all the partial fractions in the RHS of (i):

(ii) \( P(x) = [K_{1}(x - r_{2})(x - r_{3})...(x - r_{n})] + \) \( [K_{2}(x - r_{1})(x - r_{3})...(x - r_{n})] + [K_{3}(x - r_{1})(x - r_{2})...(x - r_{n})] + ...\)

We solve for \( K_{i} \) by substituting \( r_{i} \). Substitution of \( x = r_{1} \) gives:

(iii) \( P(r_{1}) = K_{1}(x - r_{2})(x - r_{3})(x - r_{4})...(x - r_{n}) = \) \( K_{1}\lim_{x\rightarrow r_{1}} \frac{Q(x)}{x-r_{1}} \)

Since \( r_{1} \) is a root of \( Q(x) \), \( \frac{Q(x)}{x-r_{1}} \) gives us an indeterminate form 0/0 by direct substitution \( x = r_{1} \) in the limit above. This suggests we can apply L’Hôpital’s rule to evaluate the limit.

(iv) \( K_{1}\lim_{x\rightarrow r_{1}} \frac{Q(x)}{x-r_{1}} = K_{1}\lim_{x\rightarrow r_{1}} \frac{Q'(x)}{1} = K_{1}Q'(r_{1}) \)

Therefore, (iii) evaluates to

(v) \( P(r_{1}) = K_{1}Q'(r_{1}) \)

(iv) \( K_{1} = \frac{P(r_{1})}{Q'(r_{1})} \)  (Solving for \( K_{1} \))

Similarly, we can apply this to any one of the numerators and obtain \( K_{n} = \frac{P(r_{n})}{Q'(r_{n})} \). This leads to the following decomposition of the rational function \( f(x) = \frac{P(x)}{Q(x)} \):

To use this method, we must be able to solve for the roots of \( Q(x) \). There are also some limitations to this method. First, all roots of \(Q(x)\) have to be unique. If \(Q(x)\) has double roots, the method does not work. If \(Q(x)\) has no roots, it does not work, such as \( \frac{2x^{2}}{x^{4}+1} \) which has a denominator that cannot be factored.

EXAMPLES

Example 1

Integrate the following rational function: \( f(x) = \frac{2x-3}{x^{2} + 2x - 3} \).

In this case, \( P(x) = 2x - 3 \), \( Q(x) = x^{2} + 2x - 3 \), and \( Q'(x) = 2x + 2 \). The roots of \(Q(x)\) are 1 and -3 since \( (x-1)(x+3) = x^{2} + 2x - 3 \).

The decomposition is:

(i) \( \frac{P(x)}{Q(x)} = \frac{P(r_{1})}{Q'(r_{1})(x-r_{1})} + \frac{P(r_{2})}{Q'(r_{2})(x-r_{2})} \)

(ii) \( \frac{2x-3}{x^{2} + 2x - 3} = \frac{P(1)}{Q'(1)(x-1)} + \frac{P(-3)}{Q'(-3)(x-(-3))} \)

(iii) \( \frac{2x-3}{x^{2} + 2x - 3} = \frac{-1}{4(x-1)} + \frac{-9}{-4(x+3)} \)

(iv) \( \frac{2x-3}{x^{2} + 2x - 3} = \frac{9}{4(x+3)} - \frac{1}{4(x-1)} \)

(v) \( \int \frac{2x-3}{x^{2} + 2x - 3} \, dx = \int \frac{9}{4(x+3)} \, dx - \int \frac{1}{4(x-1)} \, dx \)

(vi) \( \int \frac{2x-3}{x^{2} + 2x - 3} \, dx = \frac{9}{4}\ln(x+3) - \frac{1}{4}\ln(x-1) +C \)

You can check the answer by taking the derivative, which is simple for logarithmic functions; however, (iv) can be combined to show that it equals our original rational function: \( \frac{9}{4(x+3)} - \frac{1}{4(x-1)} = \) \( \frac{9(x-1) - (x+3)}{4(x+3)(x-1)} = \) \( \frac{8x-12}{4(x+3)(x-1)} = \) \( \frac{2x-3}{(x+3)(x-1)} \), which is our original rational function with the denominator factored.

The following example is from the lamar.edu website. They have done the problem using manual calculations. We will do this problem using our method.

Example 2

Integrate the rational function \( \frac{x^{2}+4}{3x^{3}+4x^{2}-4x} \).

When \( Q(x) \) does not have a contant, we have 0 as a root. In this case, one of the terms is \( \frac{K_{1}}{x} \). When x factored out, we have \( Q(x) = x(3x^{2} + 4x - 4) \). This further factors as \( Q(x) = x(x+2)(3x-2) \). The roots are \( x = 0, \, -2, \, \frac{2}{3} \).

We have \(Q'(x) = 9x^{2} + 8x - 4 \). Let’s make our substitutions. Since the degree of \(Q(x)\) is 3, we will have three terms.

(i) \( \frac{P(x)}{Q(x)} = \frac{P(r_{1})}{Q'(r_{1})(x-r_{1})} + \frac{P(r_{2})}{Q'(r_{2})(x-r_{2})} + \frac{P(r_{3})}{Q'(r_{3})(x-r_{3})} \)

(ii) \( \frac{x^{2}+4}{3x^{3}+4x^{2}-4x} = \frac{P(0)}{Q'(0)(x-0)} + \frac{P(-2)}{Q'(-2)(x-(-2))} + \frac{P(2/3)}{Q'(2/3)(x-2/3)} \)

(iii) \( \frac{x^{2}+4}{3x^{3}+4x^{2}-4x} = \frac{4}{-4x} + \frac{8}{16(x+2)} + \frac{40/9}{(16/3)(x-2/3)} \)

(iv) \( \frac{x^{2}+4}{3x^{3}+4x^{2}-4x} = -\frac{1}{x} + \frac{1}{2(x+2)} + \frac{5}{2(3x-2)} \)

(v) \( \int \frac{x^{2}+4}{3x^{3}+4x^{2}-4x} \, dx = \int -\frac{1}{x} \,dx + \int \frac{1}{2(x+2)} \,dx + \int \frac{5}{2(3x-2)} \, dx \)

(vi) \( \int \frac{x^{2}+4}{3x^{3}+4x^{2}-4x} \, dx = -\ln x + \frac{1}{2}\ln (x+2) + \frac{5}{6}\ln (3x-2) + c \)

We got the same answer using a more streamlined method.

Example 3

If the rational function has the same degree in the numerator and denominator, we can use long division to make to simpify it. Let’s decompose \( \frac{x^{2}+3x+1}{x^{2}-4} \).

Using long division:

\( \begin{array}{r}1\\ x^{2}-4\enclose{longdiv}{x^{2}+3x+1}\\x^{2} + 0x -4\\ \hline 3x+5\end{array} \)

(i) \( \frac{x^{2}+3x+1}{x^{2}-4} = 1 + \frac{3x+5}{x^{2}-4} \)

Now, let’s decompose the rational function \( \frac{3x+5}{(x+2)(x-2)} \). We have \(P(x) = 3x+5 \), \( Q'(x) = 2x \). There will be 2 terms.

(ii) \( \frac{3x+5}{(x+2)(x-2)} = \frac{P(2)}{Q'(2)(x-2)} + \frac{P(-2)}{Q'(-2)(x+2)} \)

(iii) \( \frac{3x+5}{(x+2)(x-2)} = \frac{11}{4(x-2)} + \frac{-1}{-4(x+2)} \)

(iv) \( \frac{3x+5}{(x+2)(x-2)} = \frac{11}{4(x-2)} + \frac{1}{4(x+2)} \)

Combining it all, we have

(v) \( \int \frac{3x+5}{(x+2)(x-2)} \, dx = \int 1 + \frac{11}{4(x-2)} + \frac{1}{4(x+2)} \, dx \)

(vi) \( \int \frac{3x+5}{(x+2)(x-2)} \, dx = x + \frac{11}{4}\ln(x-2) + \frac{1}{4}\ln(x+2) + c \)

THE COMPLEX NUMBERS LINK

Some functions do not factor if they do not have real roots. However, we can still use complex roots to decompose the fraction. For example, the simple rational function \( \frac{1}{x^{2} +1} \) is well-known because it is the derivative of \( \arctan x \).

We would have to evaluate complex values of the transendental functions that result from integration. I have a reference here that can be used for this purpose: i and Transcendental functions.

Example 4

The complex roots of \( \frac{1}{x^{2}+4} \) are \( 2i \) and \( -2i \). Let’s decompose this using these roots. The numerator \( P(x) \) is simply 1 and \( Q'(x) = 2x \). The degree is 2, so we will have 2 terms.

(i) \( \frac{P(x)}{Q(x)} = \frac{P(r_{1})}{Q'(r_{1})(x-r_{1})} + \frac{P(r_{2})}{Q'(r_{2})(x-r_{2})} \)

(ii) \( \frac{1}{x^{2}+4} = \frac{P(2i)}{Q'(2i)(x-2i)} + \frac{P(-2i)}{Q'(-2i)(x-(-2i))} \)

(iii) \( \frac{1}{x^{2}+4} = \frac{1}{4i(x-2i)} + \frac{1}{-4i(x+2i)} \)

(iv) \( \int \frac{1}{x^{2}+4} \, dx = \int \frac{1}{4i(x-2i)} \, dx - \int \frac{1}{4i(x+2i)} \,dx \)

(v) \( \int \frac{1}{x^{2}+4} \, dx = \frac{1}{4i}\ln(x-2i) - \frac{1}{4i}\ln(x+2i) + c\)

The logarithm of complex numbers \( \ln(a+bi) \) can be written in \( a+bi\) form as \( \dfrac{1}{2}\ln(a^2+b^2) + i\cdot \arctan\dfrac{b}{a} \). This is covered here: i and Transcendental functions. Therefore, (v) can be written as:

(vi) \( \int \frac{1}{x^{2}+4} \, dx = \frac{1}{8i}\ln(x^{2} + 4) + \frac{i}{4i}\arctan(\frac{-2}{x}) - \frac{1}{8i}\ln(x^{2} + 4) - \frac{i}{4i}\arctan(\frac{2}{x}) + c\)

(vii) \( \int \frac{1}{x^{2}+4} \, dx = -\frac{1}{4}\arctan(\frac{2}{x}) - \frac{1}{4}\arctan(\frac{2}{x}) + c\)

(viii) \( \int \frac{1}{x^{2}+4} \, dx = -\frac{1}{2}\arctan\frac{2}{x} + c\)

In (viii), the derivative of \( -\frac{1}{2}\arctan\frac{2}{x} \) is in fact the same as the derivative of \( \frac{1}{2}\arctan\frac{x}{2} \), despite their graphs are different.

Knowing the identity \( \ln(a+bi) = \dfrac{1}{2}\ln(a^2+b^2) + i\cdot \arctan\dfrac{b}{a} \), we can integrate a bit more complex functions with complex roots. In the next example, we will integrate \( \frac{1}{x^3-3x^2+4x-12} \).

Example 5

Integrate the rational function \( \frac{1}{x^3-3x^2+4x-12} \).

This function can be factored as \( \frac{1}{(x-3)(x^{2}+4)} \). When considering complex roots, it is factored as \( \frac{1}{(x-3)(x-2i)(x+2i)} \).

The numerator is 1, so \( P(x) = 1 \), \( Q'(x) = 3x^{2} - 6x + 4 \). With this much, we can start substituting into our formula to decompose the rational function. There are 3 roots so we will have three terms.

Let’s first resolve all the values. Since \( P(x) = 1 \), all of the values will be 1 in the numerator. Here are the other values we need:

\( Q'(3) = 3(3)^2 - 6(3) + 4 = 27-18+4 = 13 \)

\( Q'(2i) = 3(2i)^2 - 6(2i) + 4 = -12 - 12i + 4 = -8 - 12i \)

\( Q'(-2i) = 3(-2i)^{2} - 6(-2i) + 4 = -12 + 12i + 4 = -8 + 12i \).

(i) \( \frac{P(x)}{Q(x)} = \frac{P(r_{1})}{Q'(r_{1})(x-r_{1})} + \frac{P(r_{2})}{Q'(r_{2})(x-r_{2})} + \frac{P(r_{3})}{Q'(r_{3})(x-r_{3})} \)

(ii) \( \frac{1}{x^3-3x^2+4x-12} = \frac{1}{13(x-3)} + \frac{1}{(-8-12i)(x-2i)} + \frac{1}{(-8+12i)(x-(-2i))} \)

(iii) \( \int \frac{1}{x^3-3x^2+4x-12} \, dx = \) \( \int \frac{1}{13(x-3)} - \frac{1}{(8+12i)(x-2i)} + \frac{1}{(-8+12i)(x+2i)} \, dx \)  (Simplify the signs and integrate.)

(iv) \( \int \frac{1}{x^3-3x^2+4x-12} \, dx = \) \( \frac{1}{13}\ln(x-3) - \frac{1}{8+12i}\ln(x-2i) + \frac{1}{-8+12i}\ln(x+2i) + c \)

We will convert the logarithm of the complex number in (iv) using the identity above. Let’s work with the right side alone.

(v) \( \frac{1}{13}\ln(x-3) - \frac{1}{2(8+12i)}\ln(x^{2}+4) - \frac{i}{8+12i}\arctan(\frac{-2}{x}) + \) \( \frac{1}{2(-8+12i)}\ln(x^{2}+4) + \frac{i}{-8+12i}\arctan(\frac{2}{x}) \)

(vi) \( \frac{1}{13}\ln(x-3) + \left(\frac{1}{-16+24i} - \frac{1}{16+24i} \right)\ln(x^{2}+4) + \left( \frac{i}{8+12i} + \frac{i}{-8+12i}\right) \arctan(\frac{2}{x}) \)

(vii) \( \frac{1}{13}\ln(x-3) + \left(\frac{16+24 i + 16 - 24i}{-256-576} \right)\ln(x^{2}+4) + \left( \frac{-8i+12i^{2}+8i+12i^{2}}{-64-144}\right) \arctan(\frac{2}{x}) \)

(viii) \( \frac{1}{13}\ln(x-3) + \left(\frac{32}{-832} \right)\ln(x^{2}+4) + \left( \frac{-24}{-208}\right) \arctan(\frac{2}{x}) \)

(ix) \( \frac{1}{13}\ln(x-3) - \frac{1}{26}\ln(x^{2}+4) + \frac{3}{26} \arctan(\frac{2}{x}) \)

Therefore, \( \int \frac{1}{x^3-3x^2+4x-12} \, dx = \) \( \frac{1}{13}\ln(x-3) - \frac{1}{26}\ln(x^{2}+4) + \frac{3}{26} \arctan(\frac{2}{x}) + c \). I checked this is correct by taking the derivative. You can check also.

Also note that \( \arctan\frac{1}{a} - \arctan\frac{1}{b} = \arctan b - \arctan a \). So, although \( \arctan \frac{1}{x} \neq \arctan x \), taking the limits of integration will make them equal. Moreover, the derivatives of \( \arctan\frac{a}{x} \) is the same as the derivative of \( -\arctan \frac{x}{a} \). Hence, we can write the integral as \( \int \frac{1}{x^3-3x^2+4x-12} \, dx = \) \( \frac{1}{13}\ln(x-3) - \frac{1}{26}\ln(x^{2}+4) - \frac{3}{26} \arctan(\frac{x}{2}) + c \).

Example 6

Next, let’s try a rational function where \( P(x) \) is not 1. Let’s try \( \frac{x^{2} - 4x + 2}{x^{3} - 2x^{2}+16x-32} \). This function factors as follows: \( \frac{x^{2} - 4x + 2}{(x-2)(x+4i)(x-4i)} \). The roots are 2, 4i, and -4i. Here are all the values we need:

\( P(2) = 2^{2}-4(2)+2 = 4-8+2 = -2 \)

\( P(4i) = (4i)^{2} - 4(4i) + 2 = -16 - 16i + 2 = -14-16i \)

\( P(-4i) = (-4i)^{2} - 4(-4i) + 2 = -16 + 16i + 2 = -14+16i \)

\(Q'(x) = 3x^{2} - 4x + 16 \)

\( Q'(2) = 3(2)^{2} - 4(2) + 16 =12-8+16 = 20 \)

\( Q'(4i) = 3(4i)^{2} - 4(4i) + 16 = -48 - 16i + 16 = -32 - 16i \)

\( Q'(-4i) = 3(-4i)^{2} - 4(-4i) + 16 = -48 + 16i + 16 = -32 + 16i \)

Since we have three roots, we will have 3 terms.

(i) \( \frac{P(x)}{Q(x)} = \frac{P(r_{1})}{Q'(r_{1})(x-r_{1})} + \frac{P(r_{2})}{Q'(r_{2})(x-r_{2})} + \frac{P(r_{3})}{Q'(r_{3})(x-r_{3})} \)

(ii) \( \frac{x^{2} - 4x + 2}{x^{3} - 2x^{2}+16x-32} = \frac{-2}{20(x-2)} + \frac{-14-16i}{(-32-16i)(x-4i)} + \frac{-14+16i}{(-32+16i)(x+4i)} \)

(iii) \( \int \frac{x^{2} - 4x + 2}{x^{3} - 2x^{2}+16x-32} \, dx = \int -\frac{1}{10(x-2)} + \frac{14+16i}{(32+16i)(x-4i)} + \frac{-14+16i}{(-32+16i)(x+4i)} \, dx \)

(iv) \( = -\frac{1}{10}\ln(x-2) + \frac{14+16i}{32+16i}\ln(x-4i) + \frac{-14+16i}{-32+16i}\ln(x+4i) + c \)

Now, we use the logarithm identity.

(v) \( = -\frac{1}{10}\ln(x-2) + \frac{14+16i}{64+32i}\ln(x^{2}+16) + \) \( \frac{i(14+16i)}{32+16i}\arctan(\frac{-4}{x}) + \frac{-14+16i}{-64+32i}\ln(x^{2}+16) + \) \( \frac{i(-14+16i)}{-32+16i}\arctan(\frac{4}{x}) + c \)

(vi) \( = -\frac{1}{10}\ln(x-2) + \left(\frac{14+16i}{64+32i} + \frac{-14+16i}{-64+32i} \right)\ln(x^{2}+16) + \) \( \left(-\frac{14i-16}{32+16i} + \frac{-14i-16}{-32+16i}\right)\arctan(\frac{4}{x}) + c \)

(vi) \( = -\frac{1}{10}\ln(x-2) + \frac{11}{20}\ln(x^{2}+16) + \) \( \frac{9}{20}\arctan(\frac{4}{x}) + c \)

Therefore, \( \int \frac{x^{2} - 4x + 2}{x^{3} - 2x^{2}+16x-32} \, dx = \) \( -\frac{1}{10}\ln(x-2) + \frac{11}{20}\ln(x^{2}+16) + \) \( \frac{9}{20}\arctan(\frac{4}{x}) + c \).

The result can be confirmed by taking the derivative.

Example 7

Integrate \( \frac{1}{x^{3}-8} \).

The roots of the denominator are \( 2, \, -1 + \sqrt{3}i, \, -1 - \sqrt{3}i \). The rational expression factors as \( \frac{1}{(x-1)(x-(-1+\sqrt{3}i))(x-(-1-\sqrt{3}i))} \).

\( P(x) = 1 \)

\(Q'(x) = 3x^{2} \)

\(Q'(2) = 3(2)^{2} = 12 \)

\(Q'(-1+\sqrt{3}i) = -6 - 6\sqrt{3}i \)

\(Q'(-1-\sqrt{3}i) = -6 + 6\sqrt{3}i \)

(i) \( \frac{P(x)}{Q(x)} = \frac{P(r_{1})}{Q'(r_{1})(x-r_{1})} + \frac{P(r_{2})}{Q'(r_{2})(x-r_{2})} + \frac{P(r_{3})}{Q'(r_{3})(x-r_{3})} \)

(ii) \( \frac{1}{x^{3}-8} = \frac{1}{12(x-2)} + \frac{1}{(-6-6\sqrt{3}i)(x-(-1+\sqrt{3}i))} + \frac{1}{(-6+6\sqrt{3}i)(x-(-1-\sqrt{3}i))} \)

(iii) \( \int \frac{1}{x^{3}-8} \, dx = \int \frac{1}{12(x-2)} - \frac{1}{(6+6\sqrt{3}i)(x+1-\sqrt{3}i)} + \frac{1}{(-6+6\sqrt{3}i)(x+1+\sqrt{3}i)} \, dx\)

(iv) \( \int \frac{1}{x^{3}-8} \, dx = \frac{1}{12}\ln(x-2) - \frac{1}{6+6\sqrt{3}i}\ln(x+1-\sqrt{3}i) + \) \( \frac{1}{-6+6\sqrt{3}i}\ln(x+1+\sqrt{3}i) + c\)

(v) \( = \frac{1}{12}\ln(x-2) - \frac{1}{6+6\sqrt{3}i}\ln(x+1-\sqrt{3}i) + \) \( \frac{1}{-6+6\sqrt{3}i}\ln(x+1+\sqrt{3}i) + c\)

We will apply the logarithm of complex numbers identity. In (v), \( a = x + 1 \) and \( b = \sqrt{3} \) or \( b = -\sqrt{3} \).

(vi) \( = \frac{1}{12}\ln(x-2) - \frac{1}{12+12\sqrt{3}i}\ln((x+1)^2+(\sqrt{3})^{2}) + \frac{i}{6+6\sqrt{3}i}\arctan\frac{\sqrt{3}}{x+1} + \) \( \frac{1}{-12+12\sqrt{3}i}\ln((x+1)^2+(\sqrt{3})^{2}) + \frac{i}{-6+6\sqrt{3}i}\arctan\frac{\sqrt{3}}{x+1} + c\)

I will skip the elementary steps to combining the coefficients with i. Equation (vi) simplifies to (vii).

(vii) \( = \frac{1}{12}\ln(x-2) - \frac{1}{24}\ln(x^{2}+2x+4) + \frac{\sqrt{3}}{12}\arctan\frac{\sqrt{3}}{x+1} + c \)

Taking the derivative of (vii) does give the original rational function we started with.

If the above problem was done the traditional way, we would have to solve for A, B, and C in this setup \(\frac{1}{(x-2)(x^{2}+2x+4)} = \frac{A}{x-2} + \frac{Bx+C}{x^{2}+2x+4} \). This leads to solving the integral \( \frac{1}{12} \int \frac{1}{x-2} \, dx \, - \) \( \frac{1}{12} \int \frac{x+4}{x^{2}+2x+4} \, dx \). The integral \( \frac{1}{12} \int \frac{x+4}{x^{2}+2x+4} \, dx \) is solved by completing the squares and a u substitution. Considering that, it seems using the complex numbers link was the easier route.

Example

Integrate \( \frac{x^{5}}{x^{4}-1} \).

First, let’s divide since the power in the numerator is greater than the power in the denominator.

\( \begin{array}{r}x\\ x^{4}-1\enclose{longdiv}{x^{5} + 0x}\\x^{5} - x \\ \hline x\end{array} \)

(i) \( \frac{x^{5}}{x^{4}-1} = x + \frac{x}{x^{4} - 1} \, = \) \( x + \frac{x}{{(x+1)(x-1)(x+i)(x-i)}} \)

Let’s now work on the partial fraction. \( P(x) = x \) and \( Q'(x) = 4x^{3} \). The roots are 1, -1, i, and -i. The derivative values are:

\( Q'(1) = 4 \)

\( Q'(-1) = -4 \)

\( Q'(i) = -4i \)

\( Q'(-i) = 4i \)

(ii) \( \int \frac{P(r_{1})}{Q'(r_{1})(x-r_{1})} + \frac{P(r_{2})}{Q'(r_{2})(x-r_{2})} + \frac{P(r_{3})}{Q'(r_{3})(x-r_{3})} + \frac{P(r_{4})}{Q'(r_{4})(x-r_{4})} \, dx \)

(iii) \( \int \frac{1}{4(x-1)} + \frac{-1}{-4(x-(-1))} + \frac{i}{-4i(x-i)} + \frac{-i}{4i(x-(-i))} \, dx \)

(iv) \( \frac{1}{4}\int \frac{1}{x-1} + \frac{1}{x+1} - \frac{1}{x-i} - \frac{1}{x+i} \, dx \)

(v) \( \frac{1}{4} \left[\ln(x-1) + \ln(x+1) - \ln(x-i) - \ln(x+i) \right] \)

(vi) \( \frac{1}{4}\ln(x^{2}-1) - \frac{1}{4}\ln(x^{2}+1) \)

Putting it all together:

(vii) \( \int \frac{x^{5}}{x^{4}-1} \, dx = \) \( \frac{1}{2}x^{2} + \frac{1}{4}\ln(x^{2}-1) - \frac{1}{4}\ln(x^{2}+1) + c \)

Anirdesh.com Last Updated: July 19 2025, 14:07:00.