<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] There are many ways to evaluate this integral. One of them is to consider the function
$$I(n;a,b)=I=\int_0^1\frac{xdx}{(ax+b)^n}$$
for $b>0$, $0\neq a>-1$, and $1,2\neq n>0$. First we multiply the RHS by $1=a/a$:
$$I=\frac1a\int_0^1\frac{axdx}{(ax+b)^n}=\frac1a\int_0^1\frac{ax+b-b}{(ax+b)^n}dx$$
$$I=\frac1a\int_0^1\frac{dx}{(ax+b)^{n-1}}-\frac{b}a\int_0^1\frac{dx}{(ax+b)^n}$$
Then we focus on $$J(n;a,b)=J=\int_0^1\frac{dx}{(ax+b)^n}$$
Sub.: $$u=ax+b\Rightarrow \frac{du}a=dx$$
$$x=1\mapsto u=b+a\\ x=0\mapsto u=b$$
thus
$$J=\frac1a\int_{b}^{b+a}\frac{du}{u^n}=\frac1{a(1-n)}\left[(b+a)^{1-n}-b^{1-n}\right]$$
And since 
$$I(n;a,b)=\frac1aJ(n-1;a,b)-\frac{b}aJ(n;a,b)$$
We have that
$$I(n;a,b)=\frac1{a^2}\left[\frac{(b+a)^{2-n}-b^{2-n}}{2-n}-b\frac{(b+a)^{1-n}-b^{1-n}}{n-1}\right]$$
And of course your integral is given by $I(3;2,1)=1/18$. 
No partial fractions needed :)
PS: This even works for fractional values of $n$. Check it out here.
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