Same integration with 2 different answers? $$\int x(x^2+2)^4\,dx $$
When we do this integration with u substitution we get
$$\frac{(x^2+2)^5}{10}$$
as $u=x^2+2$
$du=2x\,dx$
$$\therefore \int (u+2)^4\,du = \frac{(x^2+2)^5}{10} + C$$ 
Although when we expand the fraction and then integrate the answer we get is different:
$x(x^2+2)^4=x^9+8x^7+24x^5+32x^3+16x$
$$\int x^9+8x^7+24x^5+32x^3+16x \,dx$$
we get
$$\frac {x^{10}}{10} +x^8+4x^6+8x^4+8x^2 + C$$
For a better idea of the questions, let's say the questions asks us to find the value of C when y(0)=1
Now,
$x=0$
$$\frac {0^{10}}{10} + 0^8 + 4(0)^6 + 8(0)^4 + 8(0)^2 + C = 1$$
$$\therefore C= 1$$
AND
$$\frac {(0+2)^5}{10} + C= 1$$
$$\therefore \frac {32}{10} + C = 1$$
$$\therefore C = 1 - 3.2 = -2.2$$
 A: Like mentioned in the comments this is all fixed if you remember your constant of integration.
$$\int x(x^2+2)^4\ dx= \frac{(x^2+2)^5}{10}+C$$
Note if you expand 
$$
\begin{split}
\frac{(x^2+2)^5}{10}&=\frac{1}{10}\left(x^{10}+5x^8(2)+10x^6(2^2)+10x^4(2^3)+5x^2(2^4)+2^5\right)\\
&=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2+\frac{32}{10}
\end{split}
$$
Notice the relation to your other way of computing the integral
$$
\int x(x^2+2)^4\ dx = \frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2 +C
$$
So lets call $F(x)=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2$ and $G(x)=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2+\frac{32}{10}$
then $F(x)-G(x)=-\frac{32}{10}$ a constant. All antiderivatives of a continuous function only differ by a constant.

Just for fun Let's see another one:

First lets use double angle for sine
$$
\int \cos x\sin x\ dx=\frac{1}{2}\int\sin 2x\ dx=-\frac{1}{4}\cos 2x +C
$$
Then substitutions $u=\sin x$
$$
\int \cos x\sin x\ dx=\int u\ du =\frac{u^2}{2}+C=\frac{\sin^2 x}{2}+C
$$
Then substitutions $u=\cos x$
$$
\int \cos x\sin x\ dx=\int -u\ du =\frac{-u^2}{2}+C=\frac{-\cos^2 x}{2}+C
$$
If you find the constant differences and  combine them in the right way you get the half angle formulas:
$$
\sin^2 x=\frac{1-\cos 2x}{2},\quad \cos^2 x=\frac{1+\cos 2x}{2}
$$
Note you can pretty quickly derive some funky trig identities in this way. For instance if you consider $\int \cos^3 x \sin^5 x\ dx$
A: You can check an antiderivative by differentiating.
$$\left(\frac{(x^2+2)^5}{10}\right)'=x(x^2+2)^4=x^9+8x^7+24x^5+32x^3+16x$$
and
$$\left(\frac {x^{10}}{10} +x^8+4x^6+8x^4+8x^2\right)'=x^9+8x^7+24+32x^3+16x$$
and the two expressions are indeed equivalent.

Now the long explanation.
Consider the binomial $x^2+a$ raised to some power $n$ and multiplied by $2x$.
$$2x(x^2+a)^m$$
which integrates as
$$\frac{(x^2+a)^{m+1}}{m+1}.$$
By the binomial theorem, the terms in the development of this antiderivative are
$$\frac1{m+1}\binom{m+1}kx^{2(m+1-k)}a^k.$$
On the other hand, the development of the initial integrand gives terms
$$2\binom mkx^{2(m-k)+1}a^k,$$ and after integration
$$\frac1{m-k+1}\binom mkx^{2(m-k)+2}a^k.$$ 
It is easy to see that all terms coincide, because
$$\frac1{m+1}\frac{(m+1)!}{k!(m+1-k)!}=\frac1{m-k+1}\frac{m!}{k!(m-k)!}=\frac{(m-1)!}{k!(m-k+1)!}.$$
Anyway, the first development holds for $0\le k\le m+1$, giving a constant term $\dfrac{a^m}{m+1}$, but the second for $0\le k\le m$ only, giving no constant term. But this does not matter, as two antiderivatives can differ by a constant.
