Different result for integral of $(x+1)^2$ depending on technique used

Question

So I might be completely wrong, and I surely am, but I tried solving a very simple looking integral, but depending on what I do to solve it, I get different results. Here's what I did for the first try: $$ \int(x+1)^2dx\\ t=x+1,\ dx=dt\\ \int t^2dt=\frac{t^3}{3}+C=\frac{(x+1)^3}{3}+C $$ Surely this is correct, but when I do not use substitution and just evaluate the integral in itself, it gives me this: $$ \int(x+1)^2dx=\int x^2+2x+1\ dx=\frac{x^3}{3}+x^2+x+C=\frac{(x+1)^3-1}{3}+C $$ So which one is it, or is there just something that allows both solutions to exist?

Answer 1

They are the same.

The $C$ in your second result equals the $C$ in your first result plus $1/3$.

Also note if you did a definite integral i.e. $\int_a^b$ then your first result gives $$ \left[ \frac{(b+1)^3}{3}+C \right ]-\left[ \frac{(a+1)^3}{3}+C \right ]\\ = \frac{(b+1)^3}{3}-\frac{(a+1)^3}{3}\\ $$ and the second results gives $$ \left[ \frac{(b+1)^3-1}{3}+C \right ]-\left[ \frac{(a+1)^3-1}{3}+C \right ]\\ = \frac{(b+1)^3}{3}-\frac{(a+1)^3}{3}\\ $$ i.e. the same again.

Answer 2

Don't forget about the $+C$! This is why you get this.

Answer 3

$$\frac{(x+1)^3}{3}+D=\frac{(x+1)^3-1}{3}+C$$

We have $$D= -\frac13+C$$

The solution differs by a constant (which will vanish if you differentiate it)

Answer 4

Constant when integrated over $t\ne$ Constant when integrated over $x$