integration by u-substitution

Question

U- substitution problem is as follows; $$\int (x^{3}+1)^{2}3x^{2}dx$$\ answer pretty easy; $$\dfrac{(x^{3}+1)^{3}}{3}+C$$\now question, when went to check and compare doing with and without the u-substitution got the following;

Without u-substitution(just multiply the integrand out; $$\int (3x^{8}+6x^{5}+3x^{2})dx=\dfrac{1}{3}x^{9}+x^{6}+x^{3}+C$$\

now , to check, multiply out the result of the u-substituion;

$$ \dfrac{(x^{3}+1)^{3}}{3}+C=\dfrac{1}{3}x^{9}+x^{6}+x^{3}+\dfrac{1}{3}+C$$\

so hopefully you see my question, $$\dfrac{(x^{3}+1)^{3}}{3}$$\ when multiplied out, generates an additional 1/3? Does this just get folded into the constant? What to make of it?

Answer 1

When you do the same integration integration $I=\int f(x) dx$ by two different methogs. You get $I_1(x)$ and $I_2(x)$, they may differ by constant only.$$I_1(x)-I_2(x)=\text{constant ~independent of $x$}.$$, Check that your case $$I_1(x)-I_2(x)=1$$

Answer 2

If we consider indefinite integrals, the answer is that an additional constant is absorbed into the integration constant, because it's not an extra degree of freedom in the set of antiderivatives. If we consider definite integrals, we note that $\int_0^x(t^3+1)^23t^2dt$ can be evaluated as$$[\tfrac13(t^3+1)^3]_a^x=\tfrac13((x^3+1)^3-(a^3+1)^3)$$by substitution or$$[\tfrac13t^9+t^6+t^3+C]_1^a=\tfrac13(x^9-a^9)+x^6-a^6+x^3-a^3$$by expanding the factorized integrand. You can verify these agree for any $a$.