Integrate 10/(x-1)(x^2+9) dx

Question

What is the answer? How do you do this? Partial fractions. Thanks.

Answer 1

Write 10/[(x-1)(x²+9)] as A/(x-1) + (Bx+C)/(x²+9)

 

So, A(x²+9) + (Bx+C)(x-1) = 10

 

     (A+B)x² + (-B+C)x + (9A-C) = 0x² + 0x + 10

 

Equating coefficients of like powers, we have the system of equations:  

A+B = 0

C-B = 0

 9A-C = 10

 

From the second equation, C = B.

 

Replacing C by B in the third equation, we get

A+B=0

9A-B=10

 

Adding the equations, 10A = 10  So, A=1

 

Thus, A = 1, B = -A = -1, C = B = -1

 

So, ∫[10/[(x-1)(x²+9)]dx = ∫[1/(x-1) + (-x-1)/(x²+9)]dx

 

                                      = ∫[1/(x-1) -x/(x²+9)-1/(x²+9)]dx

 

                                      = lnlx-1l - ½ln(x²+9) - (1/3)Tan^-1(x/3) + K  

                                                                                 (K = any constant)