Antiderivative by the method of bringing under the differential sign

This problem asks to find the antiderivative by the method of bringing a factor under the differential sign.

Problem

Find the antiderivative of the function:

\int (\frac{1}{\sqrt{4 - x^{2}}} + \frac{1}{3 + x} + \frac{1}{3 - x}) \cdot {(arcsin \frac{x}{2} + ln (3 + x) - ln (3 - x))}^{- 1} d x

Solution

Step 1

Denote the expression in parentheses by u(x). Note that this expression is a composite function whose derivative may turn out to be exactly the first bracket of the integrand.

Why this substitution?

The method of bringing under the differential sign works when the integrand can be written as u′(x)/u(x). Then u′(x) dx = du, and the integral reduces to the tabular one ∫ du/u = ln|u| + C. In our case, the expression in the denominator is a natural candidate for u(x), and the bracket in the numerator may turn out to be u′(x).

u (x) = arcsin \frac{x}{2} + ln (3 + x) - ln (3 - x)

Step 2

Compute the derivative u′(x). Use the chain rule:

u^{'} (x) = \frac{1}{\sqrt{4 - x^{2}}} + \frac{1}{3 + x} + \frac{1}{3 - x}

Step 3

As we can see, u′(x) coincides exactly with the bracket in the numerator of the original integral. Hence we can bring u′(x) dx under the differential sign:

What does bringing under the differential give?

The substitution u′(x) dx = du turns the complicated-looking integrand u′(x)/u(x) dx into the tabular integral du/u. This technique is called "bringing a factor under the differential sign". It is a special case of the substitution method performed orally, without explicitly writing out the new variable.

\int \frac{u^{'} (x)}{u (x)} d x = \int \frac{d u}{u}

Step 4

The integral of du/u is tabular:

\int \frac{d u}{u} = ln | u | + C

Step 5

Return to the variable x, substituting the original expression for u(x):

ln | arcsin \frac{x}{2} + ln (3 + x) - ln (3 - x) | + C

Answer

ln|arcsin(x/2) + ln(3+x) − ln(3−x)| + C.

Frequently asked questions

What is the essence of the method of bringing under the differential sign?

The method consists in representing part of the integrand as the differential of some function u(x): u′(x) dx = du. The integral then takes a simpler form in terms of u. Often this leads to a tabular integral.

How do I know what to bring under the differential?

Look for a pair "function and its derivative" in the integrand. If one part of the expression looks like the derivative of another part — this is a signal that the method applies. In our case the bracket u′(x) is the derivative of the expression in the denominator u(x).

How does this method differ from ordinary substitution?

Technically they are the same. In substitution we explicitly introduce a new variable t = u(x), compute dt = u′(x) dx and rewrite the integral. In bringing under the differential we do this orally: we immediately replace u′(x) dx with du, without writing out the variable t. The result is identical.

Why does the modulus appear in the answer?

When integrating ∫ du/u = ln|u| + C. The modulus is necessary because the logarithm is defined only for positive arguments. In our case the expression u(x) may be negative on some parts of the domain, so the answer is written with a modulus.