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\newp 13.  Note that you may divide numerator and denominator by
$x$ first if you wish.

If you do not divide by $x$ first, the form is
$${A\over x}+{B\over x-1}+{C\over (x-1)^2}+{Dx+E\over
x^2+1}+{Fx+G\over (x^2+1)^2}$$

If you do divide by $x$ first, then leave off the first term.

\endp

\newp 14.  Since $b-a=4$, and $n=8$, we have ${\Delta x\over
3}={1\over 6}$.  The
approximation is therefore
$${1\over6}\Bigg\lbrack 
f(0)+4f\biggl({1\over2}\biggr)+
2f(1)+4f\biggl({3\over 2}\biggr)+
2f(2)+4f\biggl({5\over 2}\biggr)+
2f(3)+4f\biggl({7\over 2}\biggr)+f(4)\Biggr]$$
\endp

\newp 15.
We can use the technique of partial fractions to replace the given 
fraction by a sum of terms which we hope will be easier to integrate, so
$${2x^2\over (x+1)(x^2+1)}= \frac{A}{x+1} + \frac{Bx+C}{x^2+1}$$
or
$$2x^2 = A (x^2+1) + (Bx+C)(x+1)\ .\leqno(\ast)$$
One can equate coefficients and get the system
$$\matrix{
x^2:&\hskip10pt&A + B = 2\cr
x:&\hskip10pt&B + C = 0\cr
constant:&\hskip10pt&A + C = 0\cr
}$$
From the last two equations, $A=B$ and hence $2A=2$: $A=1$
and then $B=1$ and $C=-1$.

OR, plug in $x=-1$ into $(\ast)$ to get $2(-1)^2 = A(2)$, or $A=1$
and then plug in $x=0$ to get $0 = 1 + C$ so $C=-1$ and then
plug in $x=1$ to get $2 = 1(2) + (B-1)(2)$ or $B=1$.

One can also do a judicious mixture of the two techniques.

In any case,
$\Int_0^1 {2x^2\over (x+1)(x^2+1)}\ dx = 
\Int_0^1 \frac{dx}{x+1} +\Int_0^1 \frac{x-1}{x^2+1}\ dx$.

$\Int_0^1 \frac{dx}{x+1} = \ln\vert x+1\vert\ \Bigg\vert_0^1=
\ln(2) - \ln(1) = \ln(2)$.

\pbreak

$\Int_0^1 \frac{x-1}{x^2+1}\ dx =
\Int_0^1 \frac{x}{x^2+1}\ dx - \Int_0^1 \frac{dx}{x^2+1} =
\frac12\ln \vert x^2+1\vert - \arctan(x) \Bigg\vert_0^1 = 
\frac12\ln(2)- \arctan(1) - \bigl(\ln(1) - \arctan(0)\bigr)
$.
Since $\arctan(1)=\frac{\pi}{4}$ and $\arctan(0)=0$,
$\Int_0^1 \frac{x-1}{x^2+1}\ d x=
\frac12 \ln(2) - \frac{\pi}{4}$ and so
$\Int_0^1 {2x^2\over (x+1)(x^2+1)}\ dx = \frac 32 \ln(2) - \frac{\pi}{4}$.

You can also do $\Int_0^1 \frac{x}{x^2+1}\ dx$ by substitution:
$u = x^2 + 1$, $du = 2x\ dx$, 

\noindent so
$\Int_0^1 \frac{x}{x^2+1}\ dx =
\frac12 \Int_1^2 \frac{du}{u} = 
\ln\vert u\vert \ \Bigg\vert_1^2 = \frac12 \bigl( \ln(2) - \ln(0)\bigr) 
= \frac12 \ln2$.


If you forgot $\Int_0^1 \frac{dx}{x^2+1}$ you can do a trig. substitution:
$x = \tan \theta$, $dx = \sec^2\theta\ d\theta$, so
$\Int_0^1 \frac{dx}{x^2+1} = 
\Int_0^{\pi/4} \frac{\sec^2\theta\ d\theta}{\sec^2\theta} =
\Int_0^{\pi/4}  d\theta = \frac{\pi}{4}$.
\endp

\newp 15.
Divide through by x and rearrange terms in the equation as
follows: 
$$ \frac{dy}{dx} + y(1+\frac{1}{x})=1\  $$ 
Now in standard notation we have that 
$P(x)=1+\frac{1}{x}$ and $Q(x)=1$. 
This yields 
$$\int P(x) dx = \int (1+\frac{1}{x})dx = x + \ln x$$ 
and
$$I(x)= e^{\int P(x) dx} = e^{(x + \ln x)} = e^{x}* e^{\ln x}= x
e^{x}\ .$$ 
The formula for a solution of the equation has the
following form 
$$y(x) = \frac{1}{I(x)} (\int I(x) Q(x) \; dx
+C)=\frac{1}{xe^x} \int xe^x
\cdot 1 \; dx + \frac{C}{xe^x}\ .$$
Integrate by parts:
$u=x$, $dv=e^x dx$.
Then $du = dx$ and $v=e^x$ so 
$$ \int xe^x dx = xe^x -\int e^x dx =  xe^x - e^x + C\ .$$
The solution is
$$y(x) = \frac{1}{xe^x}(xe^x -e^x + C) = 1 -
\frac{1}{x} + \frac{C}{xe^x}\ .$$

\endp
\end