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Don't think they'd ask that.

$\noindent Seems unlikely. But what do you mean by your question? E.g. something like, given $y' = x + y^2$, $y(0) = 1$, sketch (roughly) the graph of $y$ vs. $x$?$

$\noindent You did it right. You can (and should) check your answer, i.e. check that $5-x^{2} =\left(3-2x -x^{2}\right) -(-2-2x)$.$

$\noindent In your particular example it is $q(x) = 5-x^{2}$ and $Q(x) = 3-2x -x^{2}$, so $Q'(x) = -2-2x$. So try finding constants $a,b,c$ such that$ $$5-x^{2}\equiv a\left(3-2x -x^{2}\right) + b(-2-2x) + c$.$ $\noindent (You can find these similarly to how you find constants in partial...

$\noindent One way is as follows (general method when the numerator is a quadratic and denominator is square root of a quadratic). Say we want to find $\int \frac{q(x)}{\sqrt{Q(x)}}\,\mathrm{d}x$, where $Q(x)$ is a degree $2$ polynomial and $q(x)$ is as well.$ $\noindent 1. Write $q(x) =...

I used \underbrace{...}_{\text{...}}. (Replace the first "..." with the thing you want the brace to go under and the second "..." with what text you want to write under the brace.)

$\noindent E.g. By polynomial division, $\frac{x^{3}}{x^{2} + 1} = \underbrace{x}_{\text{quotient}} - \underbrace{\frac{x}{x^{2}+1}}_{\text{``remainder part''}}$, so$ $$\begin{align*}\int \frac{x^{3}}{x^{2} +1}\,\mathrm{d}x &= \int \left( x -\frac{x}{x^{2}+1}\right)\,\mathrm{d}x \\ &= \cdots...

$\noindent You also have to include the quotient part (this part is easy to integrate as it is just a polynomial). So rewrite the fraction as a polynomial (quotient) plus the remainder divided by the dividend and go from there.$

Its degree is less than the denominator's so we can use the method outlined earlier for that part.

$\noindent Use polynomial division first and the ``remainder term'' (remainder you get from doing the polynomial long division divided by the divisor, $x^2 + x +1$ here) will be amenable to the procedure outlined above. This is because the remainder has lower degree than the divisor. (The...

$\noindent Add and subtract $-2x +3$ to the numerator of the integrand, so the integral becomes $\int \left(1 + \frac{2x-3}{x^{2} -2x +3}\right)\,\mathrm{d}x = x + \int \frac{2x-3}{x^2 -2x +3}\,\mathrm{d}x$. This latter integral can be tackled using the procedure outlined above.$

$\noindent Steps to evaluating any integral like this (linear function on numerator and degree two polynomial (quadratic) in denominator or in a square root in denominator) is as follows. Let $q(x)$ be the quadratic present (it was $x -x^{2}$ in your case).$ $\noindent 1. Write the linear...

Re: International Baccalaureate Marathon $\noindent The relevant triangle's angle is the angle between $\overrightarrow{BA}$ and $\overrightarrow{BC}$ but you did the angle between $\overrightarrow{AB}$ and $\overrightarrow{BC}$ instead. Since you negated the first vector (and left the second...

$\noindent That fact follows from the factorisation$ $$a^{3} + b^{3} + c^{3} -3abc = (a+b+c)\left(a^{2} + b^{2} + c^{2} - ab -bc -ca\right).$$

It's basically an ad hoc method. You can also find the integral of the secant function more conventionally by using the "t-formulae", but you'd then need to use some trig. identities to convert the resulting answer into the answer I wrote in the earlier post. Incidentally, finding the integral...

$\noindent It's a well-known result that $\int \sec t \, \mathrm{d}t = \ln \left|\sec t + \tan t\right|$. One way to see this is to multiply top and bottom of the integrand by $\sec t + \tan t$ to make it $\int \frac{\sec^{2}t + \sec t \tan t}{\sec t + \tan t }\,\mathrm{d}t$ and observe that the...

$\noindent Probably. The answer to that is $\ln \left|x +\sqrt{x^{2} -4}\right|$, and can be shown with a trig. substitution. (Those ones they got rid of similar to the one you wrote can all be evaluated using trig. substitutions, so there's no reason students couldn't be expected to do them...

$\noindent If we substitute $u = 4-x^{2}$, then $\mathrm{d}u = -2x\,\mathrm{d}x \Rightarrow 5x \,\mathrm{d}x = -\frac{5}{2}\,\mathrm{d}u$. Hence$ $$\begin{align*}\int \frac{5x}{4-x^{2}}\,\mathrm{d}x &= -\frac{5}{2}\int \frac{\mathrm{d}u}{\sqrt{u}} \\ &= -\frac{5}{2}\times 2\sqrt{u}\\ &= -5u \\...

$\noindent By \textbf{inspection}, the answer is $-5\sqrt{4-x^{2}} + \sin^{-1}\left(\frac{x}{2}\right)+c$.$ $\noindent $\Big{(}$Split up the numerator and you're left with two standard integrals, i.e. $\int \frac{5x}{\sqrt{4-x^{2}}}\,\mathrm{d}x + \int \frac{1}{\sqrt{4-x^{2}}}\,\mathrm{d}x$. If...

Re: International Baccalaureate Marathon $\noindent We don't need then $\pm$ sign because we are only asked to find \emph{a} desirable vector (so only need one).$