Integration by Splitting into a Sum

To compute an indefinite integral, it’s often helpful to break the integrand into a sum of simpler expressions using the linearity property of integration.

Why use this method?

It allows you to rewrite a complex integral as a sum of basic integrals, each of which is easier to evaluate.

Note. Finding a suitable decomposition isn’t always obvious - but when it works, it can dramatically streamline the calculation.

A Practical Example

Let’s find the indefinite integral of the following expression:

$$ \int \frac{x}{x+1} \: dx $$

We begin by rewriting the integrand in a more manageable form.

One useful trick is to add and subtract 1 in the numerator:

$$ \int \frac{x+1 - 1}{x+1} \: dx $$

Now we simplify the expression algebraically, splitting the fraction into two terms:

$$ \int \left( \frac{x+1}{x+1} - \frac{1}{x+1} \right) \: dx $$

$$ \int \left(1 - \frac{1}{x+1} \right) \: dx $$

By the linearity of the integral, we can split this into two separate integrals:

$$ \int 1 \: dx - \int \frac{1}{x+1} \: dx $$

The antiderivative of the first term is simply \( x \):

$$ x + c - \int \frac{1}{x+1} \: dx $$

Note. To quickly verify: $$ D[x + c] = 1 $$

The antiderivative of the second term is the natural logarithm of the absolute value of \( x + 1 \):

$$ x + c - \log|x+1| $$

Note. To verify: $$ D[\log |x+1|] = \frac{1}{x+1} $$

There’s no need to add another constant - one constant of integration suffices.

We’ve now found the complete solution to the original integral:

$$ \int \frac{x}{x+1} \: dx = x - \log|x+1| + c $$

And we’re done.