Math problem integration by partial fractions

[tessa127]

Homework Statement

integrate (4x+3)/(x^2+4x+5)^2

Homework Equations

The Attempt at a Solution

I know to solve this problem you have to work with partial fractions, in the solution we were given they solve as followed

4x+3=A(x^2+4x+5)'+B

I don't know why they take the derivative of x^2+4x+5

thank you in advance!

 

[Mark44]

Homework Statement

integrate (4x+3)/(x^2+4x+5)^2

Homework Equations

The Attempt at a Solution

I know to solve this problem you have to work with partial fractions, in the solution we were given they solve as followed

4x+3=A(x^2+4x+5)'+B

I don't know why they take the derivative of x^2+4x+5

thank you in advance!

From what you show here they are not using partial fractions. What they're doing is an ordinary substitution, with ##u = x^2 + 4x + 5##. In this substitution, what is du? That's where the derivative you're asking about comes in.

 

[epenguin]

Homework Statement

integrate (4x+3)/(x^2+4x+5)^2

Homework Equations

The Attempt at a Solution

I know to solve this problem you have to work with partial fractions, in the solution we were given they solve as followed

4x+3=A(x^2+4x+5)'+B

I don't know why they take the derivative of x^2+4x+5

thank you in advance!

In order to solve the problem! I think I remember doing this myself for some recent problem on here.

You have a linear divided by a quadratic. The derivative of a quadratic is a linear. If you are lucky, i.e. in the simplest case the numerator will be the derivative of the denominator (multiplied by a number). In general however the derivative of the quadratic will be the numerator (multiplied by a number) plus a constant. So you will get to integrate something of form

C₁Q'/ Q + C₂/Q, where the C's are known constants.

The integral of the first fraction is C₁ ln Q , and the second fraction you expresse as partial fractions, and will get further ln 's .

 

[tessa127]

In order to solve the problem! I think I remember doing this myself for some recent problem on here.

You have a linear divided by a quadratic. The derivative of a quadratic is a linear. If you are lucky, i.e. in the simplest case the numerator will be the derivative of the denominator (multiplied by a number). In general however the derivative of the quadratic will be the numerator (multiplied by a number) plus a constant. So you will get to integrate something of form

C₁Q'/ Q + C₂/Q, where the C's are known constants.

The integral of the first fraction is C₁ ln Q , and the second fraction you expresse as partial fractions, and will get further ln 's .

thank you very much for the explanation, I will try the exercise with your method

 

[epenguin]

I will try the exercise with your method

I expect someone else discovered it first.

 

[SteamKing]

To be sure, you have a linear polynomial divided by a quadratic polynomial squared. There might be a partial fraction expansion, but it will probably be messy.

When the denominator is raised to a power, it's better to see if substitution can be used.

 

[epenguin]

To be sure, you have a linear polynomial divided by a quadratic polynomial squared. There might be a partial fraction expansion, but it will probably be messy.

When the denominator is raised to a power, it's better to see if substitution can be used.

Oops yes I had missed that the denominator was squared. So it seems they want to get it in the form Q'/Q² (multiplied by a number) which is -(1/Q)' .
So that bit can be integrated to -1/Q and you are left with some number multiplying 1/Q² which is a standard form to integrate.