Question

Evaluate the integral by partial fraction method. 

\( \begin{aligned} \int\frac{3 x-1}{(x-4)(x+1)(x-1)} dx \end{aligned} \)

Answer 1

When the degree of numerator is less than the degree of denominator then we have to use the partial fraction method provided that function should be a polynomial. 

Let 

\( \begin{aligned}\frac{3 x-1}{(x-4)(x+1)(x-1)} =\frac{A}{x-4}+\frac{B}{x+1}+\frac{C}{x-1} \end{aligned} .......(1) \)

After considering the A,B and C some constant,  we do cross multiplication. And we get,

\( \begin{aligned} \therefore \space 3 x-1&=A(x+1)(x-1)+B(x-4)(x-1)\\&+C(x-4)(x+1) \end{aligned} \)

In next step we have to find the value of constants by assuming any value to get another value. 

Putting x = 4 to get value of A,

\( \begin{aligned} \therefore 3(4)-1 &=A(4+1)(4-1) \\ \therefore 11 &=15 A \\ \therefore A &=\frac{11}{15} \end{aligned} \)

Putting x = -1 to get value of B,

\( \begin{aligned} \therefore3(-1)-1 &=B(-1-4)(-1-1)\\ \therefore-4 &=B(-5)(-2)\\ \therefore B &=\frac{-2}{5} \end{aligned} \)

Putting x = 1 to get value of C,

\( \begin{aligned} \therefore 3(1)-1 &=C(1-4)(1+1)\\ \therefore 2 &=C(-3)(2)\\ \therefore C &=\frac{-1}{3}\\ \end{aligned} \)

After getting the values of constant A,B and C, considering equation (1)

Hence,  the resolved partial fraction is,

\( \begin {aligned} \therefore \frac{3 x-1}{(x-4)(x+1)(x-1)}=\frac{\frac{11}{15}}{x-4}+\frac{\frac{-2}{5}}{x+1}+\frac{\frac{-1}{3}}{x-1} \end{aligned} \)

In this question we are going to use a partial fraction method