Problem 1. The anti-derivative of `1/(1-2x)` is `1/2 ln |1-2x| + C`.
Note the absolute values in the logarithm; without them, you'd be taking a logarithm
of a negative number, which is legitimate in complex analysis but not in real analysis.
Anyway, the answer thus becomes `1/2 ln 3`, which can be rewritten as
`ln(sqrt(3))`.

Problem 2. `y^2 exp(y^2 x)` and `2xy exp(y^2 x)` respectively.

Problem 3. If `dv/dx = x+y`, then `v = 1/2 x^2 + xy + C(y)`,
where `C(y)` is something which doesn't depend on `x`.
If `dv/dy = x-y`, then `v = xy - 1/2 y^2 + D(x)`, where
`D(x)` is something which doesn't depend on `y`. We want both
things to be true, so `v` has to do look like both of these expressions.
One solution is `v = 1/2 x^2 + xy - 1/2 y^2`; you can also add a constant
if you wish. (These are the only solutions).

Problem 4. This is a semi-circle. If `theta` ranged from `0`
to `2 pi` instead of just to `pi`, then it would be a full circle,
centered at `(2,0)` and with radius `1`.

Problem 5. The ratio test tells you that the series converges if `|x-2|`
is less than one, and diverges if `|x-2|` is greater than 1. The
only possibilities left are when `|x-2|` are equal to 1, i.e. when
`x` is 1 or 3. But the series becomes `1-1+1-1...` in the
first case and `1+1+1+...` in the second, and neither series converges
(e.g. by the zero test). So the series only converges when
`|x-2| < 1`, i.e. when `x` is strictly between 1 and 3.

Problem 6. Applying the change of variables `x = 2y`, `dx = 2 dy`
we get

int_0^8 1/2 dx/(1+x^2)where

Problem 7. If we just had to integrate `x^2`, the answer would be
`1/3 = 0.33.` But we have this extra term `sin(x^2)/300`.
But this integrand always fluctuates between `-1/300` and `1/300`,
and we're integrating over an interval of length 1, so we know that this part
of the integral isn't going to be any bigger than `1/300` in magnitude.
Therefore the answer is `0.33` to two decimal places.