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I have recently attempted the following question:

$$\int 2x(x^2 + 3) dx $$

When I attempted the question I first expanded out the brackets to achieve:

$$\int (2x^3 + 6x)dx$$

And then integrated to get:

$$ \frac{1}2x^4 + 3x^2 + c$$

I'm not sure if you're allowed to expand out the brackets before integrating, because in the mark scheme they gave the following solution:

$$\int 2x(x^2 + 3) dx = \int (\frac{d(x^2 + 3)}{dx})(x^2+3)dx $$

Let $u = x^2+3$

$$\int u\frac{du}{dx} dx= \int (u)du$$

$$= \frac{1}{2}u^2 + c$$

Substitute in $x^2+3$ for $u$

$$= \frac{(x^2+3)^2}{2}+c$$

I understand this method, however it appears to give me a different answer to my method.

As:$$\frac{(x^2+3)^2}{2} = \frac{1}{2}x^4+3x^2+\frac{9}{2}$$ Which is different to my original answer.

This means I must have a misunderstanding or I must have made a mistake somewhere, I hope someone can help me with this.

Thanks for your time.

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    $\begingroup$ What you did is fine, unless you were instructed to use substitution. (Note the two solutions differ by a constant.) $\endgroup$ Jan 3, 2016 at 17:02

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You are indeed allowed to do algebra inside an integral. The two answer differs by a constant value, which is fine.

If you differentiated both results, you would obtain the same expression, because the derivative of a constant value is $0$. You can in fact rewrite the scheme's solution as $$\frac{(x^2+3)^2}{2} + c = \frac{1}{2}x^4+3x^2+\frac{9}{2} + c = \frac{1}{2}x^4+3x^2 + c'$$

Since $c$ and $c'$ are arbitrary constants, the results are equal and both correct.

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