Question

evaluate the integral: (5x+1)/((2x+1)(x-1))

Answer 1

http://www.wolframalpha.com/input/?i=%285x%2B1%29%2F%28%282x%2B1%29%28x-1%29%29

Click on show steps..

Answer 2

solve it by a partial fraction method

Answer 3

\[\frac{5x+1}{(2x+1)(x-1)}=\frac{A}{2x+1}+\frac{B}{x-1}\]
\[5x+1=A(x-1)+B(2x+1)=x(A+2B)+(-A+B)=5x+1\]
\[ => A+2B=5, -A+B=1 \]
\[-A+B=1=> B=1+A \]
\[A+2B=5=> A+2(1+A)=5 => 3A+2=5=> A=1\]
\[A=1=> B=1+A=1+1=2\]
so we have
\[\int\limits_{}^{}\frac{5x+1}{(2x+1)(x-1)} dx=\int\limits_{}^{}(\frac{1}{2x+1} +\frac{2}{x-1}) dx\]
\[=\frac{1}{2}\ln|2x-1|+2 \ln|x-1|+C\]

Answer 4

=>A+2B=5,−A+B=1

−A+B=1=>B=1+A
How do u know that?

Answer 5

if you want wolfram's answer you need to play with ln|x-1|:
ln|x-1|=ln|(-1)(-x+1)|=ln|(-1)(1-x)| =ln[|-1| *|1-x|]
=ln[|1|*|1-x|]=ln(1)+ln|1-x|=0+ln|1-x|=ln|1-x|

Answer 6

right hand side must = left and side
x coefficient on left side must = x coefficient on right side
constant term on left side must=constant term on right side

Answer 7

1/2ln|2x−1|+2ln|x−1|+C where did that first 1/2 come from?

Answer 8

\[\int\limits_{}^{}\frac{1}{2x+1} dx\]
\[u=2x+1=> du =2 dx => \frac{1}{2} du=dx\]
\[\int\limits_{}^{}\frac{1}{2x+1} dx=\int\limits_{}^{}\frac{1}{2} \frac{du}{u}=\frac{1}{2} \ln|u|+C\]

Answer 9

oh ok, yeah i guess im just still lost on the partial fractions...i dont quite understand how we ended up with 1 and 2

Answer 10

we got two equations with 2 unknowns and solved one of the equations for B and plugged into the other equation and solved for A

Answer 11

oh! ok that makes sense!