Integration by Parts for Logarithmic and Inverse Trigonometric Functions

Why Multiply by 1?

Some functions do not have an obvious antiderivative when they appear alone. A useful strategy is to regard the function as a product with \(1\). For example, \(\ln x\) can be written as \((\ln x)(1)\), and \(\arctan x\) can be written as \((\arctan x)(1)\). That form allows us to use integration by parts.

The choice is effective because differentiating a logarithmic or inverse trigonometric function often produces a simpler expression, while the factor \(1\) has the simple antiderivative \(x\). The resulting integral is then manageable using substitution or a familiar derivative.

The Rule and the Setup

Integration by parts follows from the product rule. In its indefinite-integral form, if \(u\) and \(v\) are differentiable functions, then

For an integral such as \(\int \ln x\,dx\), the factor choices are \(u=\ln x\) and \(dv=dx\). Thus \(du=\frac{1}{x}\,dx\) and \(v=x\). The key is not just applying the formula: it is choosing factors so the new integral is easier than the original one.

For these examples, \(dv=dx\), so \(v=x\). The resulting pattern is

provided \(f\) is differentiable on the interval considered. This is the integration-by-parts formula with \(u=f(x)\) and \(dv=dx\). It is not a shortcut that automatically makes every integral easy; the remaining integral still has to be evaluated.

Worked Example: The Natural Logarithm

The product rule is what makes the method work: differentiating \(x\ln x\) produces both \(\ln x\) and \(1\). Subtracting the antiderivative of the extra \(1\) leaves precisely the desired function.

Worked Example: The Inverse Tangent

Worked Example: The Inverse Sine

Why the Method Works

Here is the product-rule justification in full. Suppose \(u\) and \(v\) are differentiable on an interval. The product rule states

Rearrange to isolate one product of a function and a derivative:

Integrating both sides gives

In differential notation, this is \(\int u\,dv=uv-\int v\,du\). For a logarithm or inverse trigonometric function by itself, choose that function as \(u\), and choose \(dv=dx\). Then \(v=x\), so the new integral contains \(x\) times the derivative of the original function. In the examples, this new integral is simpler and can be completed directly.

Choosing Factors and Checking Domains

For this particular family of integrals, the useful setup is consistent: set the logarithmic or inverse trigonometric function equal to \(u\), and set \(dv=dx\). Reversing that choice is not useful because it would require integrating the logarithmic or inverse trigonometric function in order to obtain \(v\), which is the original problem.

The function’s domain also determines where the result applies. For example, \(\ln x\) requires \(x>0\). On any interval that does not contain zero, one can also integrate \(\ln|x|\): its derivative is \(1/x\), so the same calculation gives

This formula applies separately on intervals \(x<0\) and \(x>0\); \(\ln|x|\) is not defined at \(x=0\). For \(\arcsin x\), keep the derivative calculation inside \((-1,1)\). At the endpoints, the derivative formula used in the example is not defined.

Common Pitfall: Confusing a Derivative with an Antiderivative

A frequent error is to use the derivative of a logarithm as though it were its antiderivative. For instance, the calculation

is incorrect: \(1/x\) is the derivative of \(\ln x\), not an antiderivative of \(\ln x\). Checking by differentiation exposes the problem:

The correction is to use integration by parts, which gives \(x\ln x-x+C\). Its derivative is \(\ln x\), as verified in the worked example.

A second sign-related risk appears when the remaining integral is itself negative, as in the \(\arcsin x\) example. There, \(\int \frac{x}{\sqrt{1-x^2}}\,dx=-\sqrt{1-x^2}\), and the parts formula subtracts that whole result. Keeping the parentheses makes the sign clear: \(-(-\sqrt{1-x^2})=+\sqrt{1-x^2}\).