Partial fractions - integral?

In order to integrate this by partial fractions, I would break up ((1 - 3x) / (x * (1 + x^2))) into

(1/x) - ((x+3) / (x^2 + 1)) and then attempt to integrate these separately.

We know that ∫ dx / x = ln(x), so all that we have to concentrate on is the second fraction...<<<...First result

For the moment only I am going to forget about the minus sign and I am going to break this fraction into two separate fractions.

(x / (x^2 + 1)) and (3 / x^2 + 1)

First let's look at (x / x^2 + 1).

Let u = x^2 + 1

Then du = 2dx and du/2 = dx

So this integral becomes ∫ (du/2) / u = (1/2) * ∫ du/u = 1/2 * ln(u)

Since u = x^2 + 1 we now know that:

∫ x / (x^2 + 1) dx = (ln(x^2 + 1) / 2) ...<<<...Second result and I will now put the minus sign back in here to get:

-(ln(x^2 + 1) / 2)

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Now we have to solve ∫ -3 dx / (x^2 + 1)

We know that the derivative of arctan(z) = 1 / (z^2 + 1)

I will put that proof at the bottom of this answer.

Hence ∫ (1 / (x^2 + 1)) = arctan(x) and ∫ (-3 / (x^2 + 1)) = -3 arctan(x)

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Putting all of this together we have the final answer:

ln(x) - [ (ln(x^2 + 1) / 2)] - 3 arctan(x) + C .....<<<<<.....Answer

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The Derivative of the Arctangent of x

Let y = arctan(x)

Then x = tan(y)............Differentiate both sides of this equation in terms of x. This is implicit differentiation.

d(x)/dx = d(tan(y))/dy * dy/dx

1 = d(tan(y)/dx) * dy/dx

dy/dx = 1 / d(tan(y)/dy

Since the derivative of tan(z) = sec^2(z)

dy/dx = 1 / sec^2(y)

Trig Identity: sec^2(z) = tan^2(z) + 1 for any real number z.

Note: This can be easily derived from the basic identity sin^2(x) + cos^2(x) = 1, simply by dividing both sides of this equation by cos^2(x).

So dy/dx = (1 / (1 + tan^2(y))

But looking above, x = tan(y), so:

1 + tan^2(y) = (1 / 1 + x^2) and

dy/dx = 1 / (1 + x^2)

End proof.

Decompose this expression into partial fractions by comparing coefficients: (x + 1) / (x³ - x² + x - 1) = (x + 1) / [(x - 1)(x² + 1)] (x + 1) / (x³ - x² + x - 1) = A / (x - 1) + (Bx + C) / (x² + 1) x + 1 = A(x² + 1) + Bx(x - 1) + C(x - 1) x + 1 = Ax² + A + Bx² - Bx + Cx - C x + 1 = (A + B)x² - (B - C)x + (A - C) A + B = 0 B - C = -1 A - C = 1 Solving these equations gives A = 1, B = -1, C = 0. (x + 1) / (x³ - x² + x - 1) = 1 / (x - 1) - x / (x² + 1) Integrate the expression term by term using this result: ∫ (x + 1) / (x³ - x² + x - 1) dx = ∫ [1 / (x - 1) - x / (x² + 1)] dx ∫ (x + 1) / (x³ - x² + x - 1) dx = ∫ 1 / (x - 1) dx - ∫ 2x / (x² + 1) dx / 2 ∫ (x + 1) / (x³ - x² + x - 1) dx = ln|x - 1| - ln(x² + 1) / 2 + C

Let

1-3x=A(x^2+1)+x(Bx+C) identically

where A,B & C are constants, then

when x=0, get A=1

when x=i (i^2=-1 is the imaginary unit), get

1-3i=i(Bi+C)=>

1-3i=-B+Ci=>

B=1 & C=-3

So, (1-3x)/[x(1+x^2)]=1/x-(x+3)/(x^2+1)

I would not use partial fraction method. I would just break it up.

(1 - 3x) / (x(1+x^2))

= 1/(x*(1+x^2)) - 3/(1+x^2)

1/(x(x^2+1))=

A/x+(Bx+C)/(x^2+1)

Meaning:

1=A(x^2+1)+(Bx+C)x

1=Ax^2+A+Bx^2+Cx

A=1, leaving:

0=x^2+Bx^2+Cx

B=-1:

0=Cx

C=0, leaving:

1/(x(1+x^2))=

1/x - x/(x^2+1)

Think you can take it from here?