TITLE: What did I do wrong while trying to solve this integral?
QUESTION [6 upvotes]: 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

REPLY [2 votes]: 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 ...