Integration by Partial Fractions Explained Simply

Partial fractions decomposition helps us break down complex rational expressions into simpler ones.For example, we can split this fraction into a sum of simpler fractions.There are three main types of factors we encounter in partial fractions decomposition.For linear factors like x plus a, we use a single term with constant numerator.When we have repeated linear factors, we need terms for each power up to the highest power.For quadratic factors that cannot be factored, we use a linear numerator over the quadratic denominator.Let's look at a more complex example that combines different types of factors.Remember these key points about partial fractions decomposition.In the next section, we'll learn how to solve for the coefficients in these decompositions.To solve a partial fractions problem, we start by multiplying both sides by the denominator.This clears the fractions and gives us a standard algebraic equation.Next, we distribute the terms on the right side of the equation.Now we group like terms to prepare for coefficient matching.By matching coefficients of like terms, we create a system of equations.Let's solve this system step by step to find our coefficients A and B.Adding the equations gives us two A equals five, so A equals five halves.Substituting back, we find B equals negative one half.Now we can write our complete partial fraction decomposition.Now that we have our partial fractions, let's learn how to integrate each term.There are three main types of terms we encounter when integrating partial fractions.For linear terms like A over x plus a, the integral gives us A times the natural log of the absolute value of x plus a.For terms with powers in the denominator, we use the power rule of integration, which gives us negative B over n minus 1 times the denominator raised to n minus 1.For quadratic terms, the integration depends on whether we can complete the square. If four q is greater than p squared, we use arctangent. Otherwise, we need to factor further.Let's solve our example step by step. First, we integrate two over x to get two natural log of absolute x.Next, we integrate one over x minus one to get natural log of absolute x minus one.Then, we integrate negative three over x plus one to get negative three natural log of absolute x plus one.Finally, we combine all terms and add our constant of integration C.