# 009B Sample Final 3, Problem 6

Find the following integrals

(a)  ${\displaystyle \int {\frac {3x-1}{2x^{2}-x}}~dx}$

(b)  ${\displaystyle \int {\frac {\sqrt {x+1}}{x}}~dx}$

Foundations:
Through partial fraction decomposition, we can write the fraction
${\displaystyle {\frac {1}{(x+1)(x+2)}}={\frac {A}{x+1}}+{\frac {B}{x+2}}}$
for some constants ${\displaystyle A,B.}$

Solution:

(a)

Step 1:
First, we factor the denominator to get
${\displaystyle \int {\frac {3x-1}{2x^{2}-x}}~dx=\int {\frac {3x-1}{x(2x-1)}}.}$
We use the method of partial fraction decomposition.
We let
${\displaystyle {\frac {3x-1}{x(2x-1)}}={\frac {A}{x}}+{\frac {B}{2x-1}}.}$
If we multiply both sides of this equation by  ${\displaystyle x(2x-1),}$  we get
${\displaystyle 3x-1=A(2x-1)+Bx.}$
Step 2:
Now, if we let  ${\displaystyle x=0,}$  we get  ${\displaystyle A=1.}$
If we let  ${\displaystyle x={\frac {1}{2}},}$  we get  ${\displaystyle B=1.}$
Therefore,
${\displaystyle {\frac {3x-1}{x(2x-1)}}={\frac {1}{x}}+{\frac {1}{2x-1}}.}$
Step 3:
Now, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {3x-1}{2x^{2}-x}}~dx}&=&\displaystyle {\int {\frac {1}{x}}+{\frac {1}{2x-1}}~dx}\\&&\\&=&\displaystyle {\int {\frac {1}{x}}~dx+\int {\frac {1}{2x-1}}~dx}\\&&\\&=&\displaystyle {\ln |x|+\int {\frac {1}{2x-1}}~dx.}\end{array}}}$
Now, we use  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u=2x-1.}$
Then,  ${\displaystyle du=2dx}$  and  ${\displaystyle {\frac {du}{2}}=dx.}$
Hence, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {3x-1}{2x^{2}-x}}~dx}&=&\displaystyle {\ln |x|+{\frac {1}{2}}\int {\frac {1}{u}}~du}\\&&\\&=&\displaystyle {\ln |x|+{\frac {1}{2}}\ln |u|+C}\\&&\\&=&\displaystyle {\ln |x|+{\frac {1}{2}}\ln |2x-1|+C.}\end{array}}}$

(b)

Step 1:
We begin by using  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u={\sqrt {x+1}}.}$
Then,  ${\displaystyle u^{2}=x+1}$  and  ${\displaystyle x=u^{2}-1.}$
Also, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {du}&=&\displaystyle {{\frac {1}{2}}(x+1)^{\frac {-1}{2}}dx}\\&&\\&=&\displaystyle {{\frac {1}{2{\sqrt {x+1}}}}dx}\\&&\\&=&\displaystyle {{\frac {1}{2u}}dx.}\end{array}}}$
Hence,
${\displaystyle dx=2u~du.}$
Using all this information, we get
${\displaystyle \int {\frac {\sqrt {x+1}}{x}}~dx=\int {\frac {2u^{2}}{u^{2}-1}}~du.}$
Step 2:
Now, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {\sqrt {x+1}}{x}}~dx}&=&\displaystyle {\int {\frac {2u^{2}-2+2}{u^{2}-1}}~du}\\&&\\&=&\displaystyle {\int {\frac {2(u^{2}-1)}{u^{2}-1}}~du+\int {\frac {2}{u^{2}-1}}~du}\\&&\\&=&\displaystyle {\int 2~du+\int {\frac {2}{u^{2}-1}}~du}\\&&\\&=&\displaystyle {2u+\int {\frac {2}{u^{2}-1}}~du}\\&&\\&=&\displaystyle {2{\sqrt {x+1}}+\int {\frac {2}{(u-1)(u+1)}}~du.}\end{array}}}$
Step 3:
Now, for the remaining integral, we use partial fraction decomposition.
Let
${\displaystyle {\frac {2}{(x-1)(x+1)}}={\frac {A}{x+1}}+{\frac {B}{x-1}}.}$
Then, we multiply this equation by  ${\displaystyle (x-1)(x+1)}$  to get
${\displaystyle 2=A(x-1)+B(x+1).}$
If we let  ${\displaystyle x=1,}$  we get  ${\displaystyle B=1.}$
If we let  ${\displaystyle x=-1,}$  we get  ${\displaystyle A=-1.}$
Thus, we have
${\displaystyle {\frac {2}{(x-1)(x+1)}}={\frac {-1}{x+1}}+{\frac {1}{x-1}}.}$
Using this equation, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {\sqrt {x+1}}{x}}~dx}&=&\displaystyle {2{\sqrt {x+1}}+\int {\frac {-1}{(u+1)}}+{\frac {1}{u-1}}~du}\\&&\\&=&\displaystyle {2{\sqrt {x+1}}+\int {\frac {-1}{(u+1)}}~du+\int {\frac {1}{u-1}}~du.}\\\end{array}}}$
Step 4:
To complete this integral, we need to use  ${\displaystyle u}$-substitution.
For the first integral, let  ${\displaystyle t=u+1.}$
Then,  ${\displaystyle dt=du.}$
For the second integral, let  ${\displaystyle v=u-1.}$
Then,  ${\displaystyle dv=du.}$
Finally, we integrate to get
${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {\sqrt {x+1}}{x}}~dx}&=&\displaystyle {2{\sqrt {x+1}}+\int {\frac {-1}{t}}~dt+\int {\frac {1}{v}}~dv}\\&&\\&=&\displaystyle {2{\sqrt {x+1}}+\ln |t|+\ln |v|+C}\\&&\\&=&\displaystyle {2{\sqrt {x+1}}+\ln |u+1|+\ln |u-1|+C}\\&&\\&=&\displaystyle {2{\sqrt {x+1}}+\ln |{\sqrt {x+1}}+1|+\ln |{\sqrt {x+1}}-1|+C.}\end{array}}}$

(a)   ${\displaystyle \ln |x|+{\frac {1}{2}}\ln |2x-1|+C}$
(b)   ${\displaystyle 2{\sqrt {x+1}}+\ln |{\sqrt {x+1}}+1|+\ln |{\sqrt {x+1}}-1|+C}$