Evaluate Integral

[nycmath]

Solve it.

[Eigenvalue]

[tex]\int\limits_{1}^{2} \frac{x²+1}{5x³+5x²}[/tex]dx

1. Use partial fraction decomposition to decompose [tex]\frac{x²+1}{5x³+5x²}[/tex]

[tex]\frac{x²+1}{5x³+5x²}[/tex]=[tex]\frac{A}{x+1}[/tex]+[tex]\frac{B}{x}[/tex]+[tex]\frac{C}{x²}[/tex]

x²+1=5x²A+5x(x+1)B+(5x+5)C
x²+1=5Ax²+5Bx²+5Bx+5Cx+5C
C=[tex]\frac{1}{5}[/tex]
5B+5C=0
5B=-1
B=[tex]\frac{-1}{5}[/tex]
A+B=1
A=[tex]\frac{2}{5}[/tex]



[tex]\frac{x²+1}{5x³+5x²}[/tex]=[tex]\frac{2}{5(x+1)}[/tex]+[tex]\frac{-1}{5x}[/tex]+[tex]\frac{1}{5x²}[/tex]

2. Integrate the functions
[tex]\int\limits_{1}^{2}[/tex]([tex]\frac{2}{5(x+1)}[/tex]+[tex]\frac{-1}{5x}[/tex]+[tex]\frac{1}{5x²}[/tex])dx
=[[tex]\frac{-1}{5}[/tex]ln(x)-[tex]\frac{1}{5x}[/tex]+[tex]\frac{2}{5}[/tex]ln(x+1)][tex]x_{1 }[/tex]²


3. Evaluate the bounds
=([tex]\frac{1}{5}[/tex]ln(2)-[tex]\frac{1}{10}[/tex]+[tex]\frac{2}{5}[/tex]ln(3))-([tex]\frac{1}{5}[/tex]ln(1)-[tex]\frac{1}{5}[/tex]+[tex]\frac{2}{5}[/tex]ln(2))

=([tex]\frac{1}{5}[/tex]ln(2)-[tex]\frac{1}{10}[/tex]+[tex]\frac{2}{5}[/tex]ln(3))-(-[tex]\frac{1}{5}[/tex]+[tex]\frac{2}{5}[/tex]ln(2))

=[tex]\frac{2}{5}[/tex]ln(3)-[tex]\frac{3}{5}[/tex]ln(2)+[tex]\frac{1}{10}[/tex]

Note: there is no need to add a constant of integration as it is a definite integral

[nycmath]

[tex]\int\limits_{1}^{2} \frac{x²+1}{5x³+5x²}[/tex]dx

1. Use partial fraction decomposition to decompose [tex]\frac{x²+1}{5x³+5x²}[/tex]

[tex]\frac{x²+1}{5x³+5x²}[/tex]=[tex]\frac{A}{x+1}[/tex]+[tex]\frac{B}{x}[/tex]+[tex]\frac{C}{x²}[/tex]

x²+1=5x²A+5x(x+1)B+(5x+5)C
x²+1=5Ax²+5Bx²+5Bx+5Cx+5C
C=[tex]\frac{1}{5}[/tex]
5B+5C=0
5B=-1
B=[tex]\frac{-1}{5}[/tex]
A+B=1
A=[tex]\frac{2}{5}[/tex]



[tex]\frac{x²+1}{5x³+5x²}[/tex]=[tex]\frac{2}{5(x+1)}[/tex]+[tex]\frac{-1}{5x}[/tex]+[tex]\frac{1}{5x²}[/tex]

2. Integrate the functions
[tex]\int\limits_{1}^{2}[/tex]([tex]\frac{2}{5(x+1)}[/tex]+[tex]\frac{-1}{5x}[/tex]+[tex]\frac{1}{5x²}[/tex])dx
=[[tex]\frac{-1}{5}[/tex]ln(x)-[tex]\frac{1}{5x}[/tex]+[tex]\frac{2}{5}[/tex]ln(x+1)][tex]x_{1 }[/tex]²


3. Evaluate the bounds
=([tex]\frac{1}{5}[/tex]ln(2)-[tex]\frac{1}{10}[/tex]+[tex]\frac{2}{5}[/tex]ln(3))-([tex]\frac{1}{5}[/tex]ln(1)-[tex]\frac{1}{5}[/tex]+[tex]\frac{2}{5}[/tex]ln(2))

=([tex]\frac{1}{5}[/tex]ln(2)-[tex]\frac{1}{10}[/tex]+[tex]\frac{2}{5}[/tex]ln(3))-(-[tex]\frac{1}{5}[/tex]+[tex]\frac{2}{5}[/tex]ln(2))

=[tex]\frac{2}{5}[/tex]ln(3)-[tex]\frac{3}{5}[/tex]ln(2)+[tex]\frac{1}{10}[/tex]

Note: there is no need to add a constant of integration as it is a definite integral

Wow! Impressive work!