Question

Find the anti derivative of (1 + 2x)^(2)

I have got 2y^(3)/3 + 2y^(2) + y
but wolfram alpha claims:
https://www.wolframalpha.com/input/?i=integrate+%281%2B2x%29^%282%29

Where am i going wrong?

I foiled it out to

4x^(2) + 4x + 1

Answer 1

then took the anti derivative but I have no idea what I did wrong

Answer 2

also Im aware I didnt add plus c but it isn't important

Answer 3

i guess you are worng ... there is no y, and with what did you integrate
the right answer would be
https://www.wolframalpha.com/input/?i=integrate+%281%2B2x%29%5E%282%29+dx

Answer 4

yeah pretend x = y

Answer 5

first of all .. expand that square and try to integrate each term with respect to x

Answer 6

We have
\[\int (1+2x)^2 dx\]
Let
\[1+2x=t\]
then
\[2dx=dt\]
or
\[dx=\frac{dt}{2}\]
so we have now
\[\int (t)^2 \frac{dt}{2}\]
We know that
\[\int x^n=\frac{x^{n+1}}{n+1}\]
so we get
\[\int (t)^2 \frac{dt}{2}=\frac{t^3}{3 \times 2}+c\]
we know t=1+2x
so we get
\[\int (1+2x)^2 dx=\frac{(1+2x)^3}{6}+c\]

Answer 7

oh thanks ash

Answer 8

wait why isn't it 2dt = dx?

Answer 9

We have assumed
\[1+2x=t\]
so differentiate both sides we get
\[2dx=dt\]
or
\[dx=\frac{dt}{2}\]

Answer 10

I still dont get it :l

Answer 11

Okay
We assumed
\[1+2x=t\]
Let's differentiate this with respect to x
or
\[\frac{d}{dx}(1+2x)=\frac{d}{dx}(t)\]
We get
\[\frac{d}{dx}(1)+\frac{d}{dx}(2x)=\frac{d}{dx}(t)\]
We get now
\[0+2=\frac{dt}{dx}\]
Multiply both sides by dx
\[2dx=dt\]

Answer 12

Does it help?

Answer 13

ok I think I see and yes a lot