<s>[INST] Integration: U-Substitution Question:(part a)
By using the substitution, $ u=2x+7$
So here is what I did, however, halfway through I had to look at wolframalpha, and don't quite understand what they do to get the answer.

$\int$ $x$$\sqrt{2x+7}$ $dx$

$du$=$2 $ $dx$
$=\frac{1}{4}$$\int$$(u-7)$$\sqrt{u}$ $du$

$=\frac{1}{4}$$\int$$(u^3/_2-7\sqrt{u})$ $du$

$=\frac{1}{4}$$\int$$(u^3/_2 $$du$$ -\frac{7}{4})$ $\int$$\sqrt{u}$ $du$
Here where they separate the integrals, I don't understand why they do it and if it is a calculus principle I do not know, and if that's the case, can someone please explain to me? thanks.  Also this next step as they integrate, they don't integrate the fraction $\frac{7}{4}$,which reminds me, how do they even get the fraction? It must be some piece of calculus about Integration I am missing out on.
$=$$\frac{u^5/_2}{10}$$-$$\frac{7}{4}$ $\int$$\sqrt{u}$ $du$
I do, however, understand after this point, but of-course I am unsure of the method used above.
$=$$\frac{u^5/_2}{10}$$-$$\frac{7u^3/_2}{6}$ $+$ $C$
$=$$\frac{(2x+7)^5/_2}{10}$$-$$\frac{7(2x+7)^3/_2}{6}$$+$ $C$
$\therefore$$\int$ $x$$\sqrt{2x+7}$ $dx$ = $\frac{\sqrt(2x+7)^5}{10}$$-$$\frac{7\sqrt(2x+7)^3}{6}$$+$ $C$
Which indeed is the correct answer.
Can anyone help me with how wolframalpha split the integral at the middle part and also why they didn't integrate $\frac{7}{4}$.Help is greatly appreciated thank you.
(part B) - I don't have a clue on how to do this one.
It's the same integral however they make, $u^2=2x+7$
if anyone can help me with this as well please?
 [/INST] If $u^2 = 2x + 7$ then $ x = \frac{u^2 - 7}{2}$ and $dx = udu$ 
$ \int x \sqrt{2x + 7} dx \ = \ \int \frac{u^2 - 7}{2} \sqrt{u^2} u du \ = \ \frac{1}{2} \int (u^4 - 7u^2) du$
$ \frac{1}{2} [\frac{u^5}{5} - \frac{7u^3}{3} ] \ + \ C \ = \ \frac{ (2x + 7)^2 \sqrt{2x + 7} }{10} \ - \ \frac{ 7(2x + 7) \sqrt{2x + 7} }{6} \ + \ C$
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