How do you evaluate #int 1/(x+7)(x^2+9)# for [-1, 3]?

Question

Samantha Adkison · Accepted Answer

#-24+58ln(5/3)# the given #(x^2+9)/(x+7)# can be written as #x+(-7x+9)/(x+7)# which can be furthur reduced as #x-7+58/(x+7)# #int_ -1^3##x-7+58/(x+7)dx#=#int_-1^3##(x^2+9)/(x+7)##dx# #int_ -1^3##xdx-7dx+58/(x+7)dx# #x^2/2-7x+58ln(x+7)#from #[-1 3]# #9/2-7*3+58ln10#-#(1/2-7*(-1)+58ln6)# #4-28+58ln(10/6)# #-24+58ln(5/3)# # ln stands for Natural Logarithm log_e(x)#

Evelyn Agard · Answer

undefined To evaluate $ \int_{-1}^{3} \frac{1}{(x+7)(x^2+9)} $, follow these steps:

1. Use partial fraction decomposition to express the integrand as a sum of simpler fractions.
2. Integrate each term separately.
3. Evaluate the definite integral by substituting the upper limit and subtracting the result of substituting the lower limit.

Here's the breakdown:

1. Perform partial fraction decomposition on $ \frac{1}{(x+7)(x^2+9)} $:

$ \frac{1}{(x+7)(x^2+9)} = \frac{A}{x+7} + \frac{Bx + C}{x^2+9} $

2. Multiply both sides by $ (x+7)(x^2+9) $ to clear the denominators:

$ 1 = A(x^2+9) + (Bx + C)(x+7) $

3. Equate coefficients of like terms:

For constants: $ 1 = 9A + 7C $  
   For $ x $ terms: $ 0 = B + 7A $  
   For $ x^2 $ terms: $ 0 = A $

Solving these equations yields:  
   $ A = 0 $  
   $ B = -\frac{1}{16} $  
   $ C = \frac{1}{16} $

4. Rewrite the integrand:

$ \frac{1}{(x+7)(x^2+9)} = \frac{-\frac{1}{16}}{x+7} + \frac{\frac{1}{16}x}{x^2+9} + \frac{\frac{1}{16}}{x^2+9} $

5. Integrate each term:

$ \int \frac{-\frac{1}{16}}{x+7} \, dx = -\frac{1}{16} \ln|x+7| $  
   $ \int \frac{\frac{1}{16}x}{x^2+9} \, dx = \frac{1}{32} \ln|x^2+9| $  
   $ \int \frac{\frac{1}{16}}{x^2+9} \, dx = \frac{1}{48} \arctan\left(\frac{x}{3}\right) $

6. Evaluate the definite integral:

$ \left[-\frac{1}{16} \ln|x+7| + \frac{1}{32} \ln|x^2+9| + \frac{1}{48} \arctan\left(\frac{x}{3}\right)\right]_{-1}^{3} $

Substitute the upper limit:  
   $ = \left[-\frac{1}{16} \ln|3+7| + \frac{1}{32} \ln|3^2+9| + \frac{1}{48} \arctan\left(\frac{3}{3}\right)\right] $

Substitute the lower limit:  
   $ - \left[-\frac{1}{16} \ln|-1+7| + \frac{1}{32} \ln|(-1)^2+9| + \frac{1}{48} \arctan\left(\frac{-1}{3}\right)\right] $

7. Simplify the expression to find the value of the definite integral.