009B Sample Final 3, Problem 6

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Find the following integrals

(a)  $\int {\frac {3x-1}{2x^{2}-x}}~dx$ (b)  $\int {\frac {\sqrt {x+1}}{x}}~dx$ Foundations:
Through partial fraction decomposition, we can write the fraction
${\frac {1}{(x+1)(x+2)}}={\frac {A}{x+1}}+{\frac {B}{x+2}}$ for some constants $A,B.$ Solution:

(a)

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

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