<s>[INST] Integration of $\int_{}^{}\frac{1}{x(1+x)^3}dx $ we got this integration problem 
$$\int_{}^{}\frac{1}{x(1+x)^3}dx $$ 
it seems a fairly simple problem but what i am struggling with it is doing its partital fractions
$\int_{}^{}\frac{1}{x(1+x)^3}dx $ =$\int_{}^{}\frac{a}{x}dx$ +$\int_{}^{}\frac{b}{(1+x)^3}dx$+
$\int_{}^{}\frac{c}{(1+x)^2}dx$ + $\int_{}^{}\frac{d}{1+x}dx$ now how to get values of $a,b,c,d$ ? it seems confusion by doing it by $a(1+x)^3+$ $bx+$ $cx(1+x)$ $+d(x(1+x)^2) =1$ is there any other way to break it into partial fractions and solve it ?
 [/INST] Ok, your equation is right:
$$a(1 + x)^3 + bx + cx(1 + x) + d(1 + x)^2 = 1$$
Rewriting this:
$$\begin{align}
a(1 + 3x + 3x^2 + x^3) + bx + cx + cx^2 + dx(1 + 2x + x^2) & = 1 \\
a + 3ax + 3ax^2 + ax^3 + bx + cx+cx^2+dx+2dx^2+dx^3& = 1 \\ 
\end{align}$$
Grouping the terms by the powers of x:
$$\begin{align}
(a+d)x^3 + (3a+2d+c)x^2 + (3a+b+c+d)x + a = 1
\end{align}$$
Now as you can see, there are no $x^3, x^2, $ or $x$ terms on the right side of the equation. Therefore because the RHS (right hand side) has to equal the LHS (left hand side), we can make a system of equations for $a, b, c$ and $d$
$$\begin{align}
x^3: a +d= 0 \\ 
x^2: 3a + 2d+c = 0 \\
x : 3a+b+c+2d = 0 \\
Constants: a = 1
\end{align}$$
Solving this we get $a = 1,$ $b = -1,$ $c = -1,$ $d = -1$
So the partial fraction decomposition is equal to:
$$\frac{-1}{(1+x)^3} + \frac{-1}{(1+x)^2} + \frac{1}{x} + \frac{-1}{(1+x)}$$
The rest you can integrate. 
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