Unit 1 · Lesson 5 of 7

Break a rational integrand into recognizable pieces.

A rational response with multiple factors can integrate more easily when decomposed into simple reciprocal terms.

Change the input. Follow the mathematics.

What you'll be able to do

Set up the model and calculate its definite integral.
Connect the equation to the changing visual.
Explain the assumptions and identify a misleading interpretation.

Predict before exploring

What will change if you increase upper bound b? Predict the direction, then explain the response.

A model you can changeMove one control at a time.

Read the model

\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2},\quad\int_0^b=\ln\frac{2(b+1)}{b+2}

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

Rational integrand

Upper bound b2

definite integral

I=\ln\frac{2(2+1)}{2+2}=0.405465

Solve coefficient identities after multiplying by the denominator. The two logarithmic terms combine after evaluating endpoints.

Assumptions, units, and limits

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

From the situation to an equation

First ensure the rational expression is proper; otherwise perform polynomial division. Distinct linear factors need constant numerators, repeated factors need all powers, and irreducible quadratic factors need linear numerators. Solve a polynomial identity for coefficients.

\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2},\quad\int_0^b=\ln\frac{2(b+1)}{b+2}

Keep the reasoning visible

\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2}

Solve coefficient identities after multiplying by the denominator. The two logarithmic terms combine after evaluating endpoints.

What makes this model useful?

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

Decomposition must be algebraically identical on the original domain. Check poles and interval validity before integrating.

A worked example

Calculate the starting scenario

Use upper bound b = 2 .
$I=\ln\frac{2(2+1)}{2+2}=0.405465$
Solve coefficient identities after multiplying by the denominator. The two logarithmic terms combine after evaluating endpoints. The checked quantity is definite integral.

Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the definite integral using upper bound b 2 . State the meaning of its sign and units.

Checked quantity: definite integral · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Explore the visual solution

Read the model

\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2},\quad\int_0^b=\ln\frac{2(b+1)}{b+2}

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

Rational integrand

Upper bound b2

definite integral

I=\ln\frac{2(2+1)}{2+2}=0.405465

Solve coefficient identities after multiplying by the denominator. The two logarithmic terms combine after evaluating endpoints.

Assumptions, units, and limits

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2},\quad\int_0^b=\ln\frac{2(b+1)}{b+2}

Calculate and explain

Apply the idea to a new scenario

Integrate 1/[( $x+1$ )( $x+2$ )] from 0 to 4.

Checked quantity: definite integral · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Explore the visual solution

Read the model

\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2},\quad\int_0^b=\ln\frac{2(b+1)}{b+2}

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

Rational integrand

Upper bound b4

definite integral

I=\ln\frac{2(4+1)}{4+2}=0.510826

Solve coefficient identities after multiplying by the denominator. The two logarithmic terms combine after evaluating endpoints.

Assumptions, units, and limits

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2},\quad\int_0^b=\ln\frac{2(b+1)}{b+2}

Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual

Read the model

\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2},\quad\int_0^b=\ln\frac{2(b+1)}{b+2}

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

Rational integrand

Upper bound b2

definite integral

I=\ln\frac{2(2+1)}{2+2}=0.405465

Solve coefficient identities after multiplying by the denominator. The two logarithmic terms combine after evaluating endpoints.

Assumptions, units, and limits

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. First ensure the rational expression is proper; otherwise perform polynomial division. Distinct linear factors need constant numerators, repeated factors need all powers, and irreducible quadratic factors need linear numerators. Solve a polynomial identity for coefficients.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

Completion is saved on this browser. This is separate from account practice progress.