009B Sample Final 3, Problem 6

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:
Therefore, 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, $dx=2udu$.
Using all this information, we get

Step 2:

Final Answer:
(a)   $\ln |x|+{\frac {1}{2}}\ln |2x-1|+C$
(b)

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