What did I do wrong while trying to solve this integral? So the question is $$\int x^3\ln(x+1)\,dx$$ and I did it this way:
$$= {1\over 4}\ln(x+1)x^4 - {1\over 4}\int {x^4\over x+1}\,dx$$
$$= {1\over 4}\ln(x+1)x^4 - {1\over 4}\int {x^4-1+1\over x+1}\,dx$$
$$= {1\over 4}\ln(x+1)x^4 - {1\over 4}\int {(x^2 -1)(x^2+1)\over x+1}\,dx - {1\over 4}\int{1\over x+1}\,dx$$
$$= {1\over 4}\ln(x+1)x^4 - {1\over 4}\int (x -1)(x^2+1)\,dx - {1\over 4}\int{1\over x+1}\,dx$$
$$= {1\over 4}\ln(x+1)x^4 - {1\over 4}\int x^3\,dx - {1\over 4}\int x\,dx + {1\over 4}\int x^2\,dx +{1\over 4}\int dx - {1\over 4}\int {1\over x+1}\,dx$$
$$= {1\over 4}\ln(x+1)x^4 - {x^4\over 16} - {x^2\over 8} + {x^3\over 12} + {x\over 4} - {1\over 4}\ln(x+1) + c$$
I looked the solution up on another website, and although it had a similar answer, when I graphed both of the equations I found that they were not equal and a small constant of: $${25\over 48}$$ was the difference between the two equations. Is this small constant lost in c? If so is my solution still right? If not, why?
Sorry if I missed something obvious :P
 A: Your process of integrating the given function using integration by parts is absolutely correct.
Now coming to the constant of integration '$c$' - $c$ is a constant we have defined for mathematical consistency. We have to include $c$ while performing indefinite integration since there exist an infinite number of potential functions with the same derivative as the function in the integrand. So $c$ itself can take an infinite number of values.
Hence, the required factor of $25/48$ can very well be incorporated into $c$.
Hope this helped!
A: Whenever you are integrating a function $f(x)$ you actually have two options. You are either performing a definite integral like :
$$A=\int_{a}^{b} f(x)dx$$where $A$ is actually the area under the curve of $f(x)$ with the $X$ Axis from the interval $[a,b]$ . Definite integral gives you area.
If you are performing an indefinite integral like:
$$\int f(x)dx = g(x)$$
You are actually trying to find a function $g(x)$ which on differentiation will give you $f(x)$. Now the beauty of this is that not only $g(x)$ but $g(x)+1$, $g(x)+ \pi$ and all functions of the form $g(x)$ when differentiated will give the function $f(x)$. So in general the answer is 
$$\int f(x) = g(x) +c$$
which is nothing but a family of curves.
Hope this helps ...
A: If your result is
$$f(x) + c$$
and the other result is
$$\left(f(x) + \frac {25} {48}\right) +c$$
then both represent the same set of functions - you may see it by using the linear substitution $c_1 = \frac {25} {48} +c\ $ to write that other result as
$$f(x) + c_1$$
And that $c_1$ is as good as $c$, both representing an arbitrary real number.

Note:
The notation $f(x) + c$ means the set of functions:
$$\left\{f(x) + c;\  c \in R\right\} $$
