<s>[INST] Integral check. Is partial fractions the only way? I'm getting a different answer from wolfram and I have no idea where. I have to integrate:
$$\int_0^1 \frac{xdx}{(2x+1)^3}$$
Is partial fractions the only way?
So evaluating the fraction first:
$$\frac{x}{(2x+1)^3} = \frac{A}{2x+1} + \frac{B}{(2x+1)^2} + \frac{C}{(2x+1)^3}$$
$$x = A(2x+1)^2 + B(2x+1) + C$$
$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$
$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$
$$x = x^2(4A) + x(4A+2B) + A + B + C$$
$4A = 0$ and $4A+2B = 1$ and $A + B + C = 0$ so $A = 0$ and $B=\frac{1}{2}$ and $C = \frac{-1}{2}$
Is the partial fraction part right?
So then I get:
$$\int_0^1 \frac{xdx}{(2x+1)^3} = \int_0^1 \frac{dx}{(2(2x+1)^2)} - \int_0^1 \frac{dx}{2(2x+1)^3}$$
for both, using $u = 2x+1$ and $\frac{du}{dx} = 2$ and $du = 2dx$ and $\frac{du}{2} = dx$, 
$$\frac{1}{4} \int \frac{du}{u^2} - \frac{1}{4} \int \frac{du}{u^3}$$
$$ = [\frac{1}{4} - u^{-1} - \frac{1}{4} \cdot \frac{1}{-2} u^{-2} ]_1^3$$
I'm going the route of changing the limits to the new u and since $u = 2x+1$, when $x = 0, u = 1$ and when $x = 1, u = 3$. Is this the right path?
finally I get
$$[\frac{-1}{4}u^{-1} + \frac{1}{8}u^{-2} ]_1^3$$
I plug in numbers but I get a different answer than wolfram...
 [/INST] $$\int_0^1 \frac{xdx}{(2x+1)^3} = 
  \int_0^1 \frac{dx}{2(2x+1)^2} - \int_0^1 \frac{dx}{2(2x+1)^3}$$
$\displaystyle \int \dfrac{dx}{2(2x+1)^2}:$
$\qquad u=2x+1,\ x=\dfrac{u-1}{2}, \ dx = \dfrac 12 du$
$\qquad \displaystyle \int \dfrac{dx}{2(2x+1)^2} = 
                      \int \dfrac{du}{4u^2} = -\dfrac{1}{4u}$
$\qquad x=0 \mapsto u=1, \ x=1 \mapsto u=3$
$\qquad \left[-\dfrac{1}{4u} \right]_1^3 = -\dfrac{1}{12} + \dfrac 14 = \dfrac 16$
$\displaystyle \int \dfrac{dx}{2(2x+1)^3}:$
$\qquad u=2x+1,\ x=\dfrac{u-1}{2}, \ dx = \dfrac 12 du$
$\qquad \displaystyle \int \dfrac{dx}{2(2x+1)^3} = 
                      \int \dfrac{du}{4u^3} = -\dfrac{1}{8u^2}$
$\qquad x=0 \mapsto u=1, \ x=1 \mapsto u=3$
$\qquad \left[-\dfrac{1}{8u^2} \right]_1^3 = -\dfrac{1}{72} + \dfrac 18 = \dfrac 19$
$$\int_0^1 \frac{dx}{2(2x+1)^2} - \int_0^1 \frac{dx}{2(2x+1)^3}
   = \dfrac 16 - \dfrac 19 = \dfrac{1}{18}$$
There is also another way to solve for $A, B$, and $C$.
$$\frac{A}{2x+1} + \frac{B}{(2x+1)^2} + \frac{C}{(2x+1)^3} =\frac{x}{(2x+1)^3}$$
$$A(2x+1)^2 + B(2x+1) + C = x$$
Let $x = -\dfrac 12$ and you get
\begin{align}
   C &= -\dfrac 12 \\
   A(2x+1)^2 + B(2x+1) - \dfrac 12 &= x \\
   2A(2x+1)^2 + 2B(2x+1) &= 2x+1 \\
   2A(2x+1) + 2B &= 1
\end{align}
Again, let $x = -\dfrac 12$ and you get
\begin{align}
   2B &= 1 \\
   B &= \dfrac 12 \\
   2A(2x+1) + 1 &= 1 \\
   A &= 0
\end{align}
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