Integration with partial fractions...

You could use the substitution u = x + 1 and it will come out nicely. For future reference, there is a key on the bottom left of the keyboard | that does a good modulus.

Part of the question says to write it in the form A + B/(x+1) first

Maybe try dividing it using the grid method or long division, looks like if you do that you'll get your A and the remainder/(x+1), which you can then integrate.

I think that should open the question up for you!

whats the grid method? ive never heard of it before.

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You could use the substitution u = x + 1 and it will come out nicely. For future reference, there is a key on the bottom left of the keyboard | that does a good modulus.

Partial fractions aren't what you need here. Try polynomial division first.

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I used that method and I would prefer using this method however, in my final answer I keep getting -2 in the expression unless that is the constan term c?

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havent done polynomial division I think :/
how would you do that?

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havent done polynomial division I think :/
how would you do that?

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If you're doing C4, you should have been taught algebraic polynomial division, which is the easiest method I've come across of dividing polynomials.


In this question there is an easier way, let $\frac{x(2x+1)}{x+1} \equiv A + \frac{B}{x+1}$, then multiply by $x+1$ and equate coefficients.

Oh ive come across that before but why isn't the A a fraction?

If you're doing C4, you should have been taught algebraic polynomial division, which is the easiest method I've come across of dividing polynomials.


In this question there is an easier way, let $\frac{x(2x+1)}{x+1} \equiv A + \frac{B}{x+1}$, then multiply by $x+1$ and equate coefficients.

Oh ive come across that before but why isn't the A a fraction?

You cover long division of polynomials in C1 I think. Then there is an algebraic method in C4. I've never heard of the grid method.

At this stage, the easiest method is the algebraic one, where you would have one polynomial P(x) and another of lower order Q(x) and you would find the other f(x) by setting P(x) = Q(x)f(x).

So for example if you had a polynomial $x^3-4x^2-2x+3$ and wanted to divide by x+1, you would set $x^3-4x^2-2x+3 \equiv (x+1)(ax^2+bx+c)$ and expand and equate coefficients to find the values of a, b, and c.

There is a slight mistake here, that should read:
$Ax+B+\dfrac{C}{x+1}$
The $A$ is not a fraction because the order of the polynomial in the numerator exceeds that of the denominator.

There is a trick of inspection to these that I can show you if you're interested.

EDIT: I'll stick it in a spoiler below, cos I'm off to bed.

The grid method is basically backwards grid multiplication.

Say x^2+x+4 / x+3

you would set x and 3 as the left column, then write powers of x to recreate the original equation.
i.e.

.....x
x x^2
3 3x

We put x at the top to begin with because there's an x^2 in the equation and we know obviously x times x is x^2. If 3x^2 was in the equation, we would've written 3x, not x, at the top.

Now, we have 3x from the grid, but in the original equation there is just x. So we put in -2 to reduce 3x to x, because -2 times x is -2x, and -2x + 3x is x.
hence:
....x....-2
x x^2 -2x
3 3x -6

Now, we can't manipulate the grid furtherly because to change the -6 we would need to go into the negative powers of x, which aren't present in the equation.

for the answer, we simply read the top line, so x-2, and for the remainder, its whatever we need to add to the grid, which is x^2+x-6, to get the original equation, x^2+x+4. Clearly, 4--6 = 10.

Hence, x^2+x+4 / x+3 = x-2 + 10/(x+3).

It's such a quick method to use when you're used to it! Hope I've explained it well!

when you got to the point of X+1-1/X+1, why is there plus between X+1/X+1 + 1/X+1. Wouldn't it be a minus?