Integrating The Inverse Tangent Of An Affine Argument

From arctan x to arctan(ax + b)

The tutorial on the plain inverse tangent gave \(\int\arctan x\,dx=x\arctan x-\frac12\ln(1+x^2)+C\). Many problems instead contain an affine argument, such as \(\arctan(2x)\) or \(\arctan(x+1)\). The structure is the same, but the inner linear function introduces a factor from the chain rule. Two routes lead to the answer: substitute first and then use the known formula, or apply integration by parts directly with the shifted argument.

Either route needs the same ingredient, the derivative \(\frac{d}{dx}\arctan(ax+b)=\frac{a}{1+(ax+b)^2}\). The factor \(a\) in the numerator is the chain rule, and it is the source of the \(\frac1a\) that appears in the final answer.

Deriving the Formula

Let \(w=ax+b\), so \(dw=a\,dx\) and \(dx=\frac{dw}{a}\). Then \(\int\arctan(ax+b)\,dx=\frac1a\int\arctan w\,dw\). The known antiderivative of \(\arctan w\) gives

Replacing \(w\) by \(ax+b\) produces the stated result. Nothing here requires \(a>0\); the formula holds for negative \(a\) as well, because the substitution is valid for any nonzero slope and \(1+w^2\) is always positive.

Checking by Differentiation

Differentiate the bracket with respect to \(x\). The product rule gives \(a\arctan(ax+b)+(ax+b)\cdot\frac{a}{1+(ax+b)^2}\), and the logarithm contributes \(-\frac12\cdot\frac{2a(ax+b)}{1+(ax+b)^2}=-\frac{a(ax+b)}{1+(ax+b)^2}\). The rational terms cancel, leaving \(a\arctan(ax+b)\). Multiplying by the outside factor \(\frac1a\) returns \(\arctan(ax+b)\), as required.

Worked Examples

A Reliable Procedure

Common Mistakes

The most frequent error is forgetting the factor \(\frac1a\), writing \((ax+b)\arctan(ax+b)-\frac12\ln(1+(ax+b)^2)\) with no outside factor. Its derivative is \(a\arctan(ax+b)\), which is \(a\) times too large. A second error is to simplify \(\frac1a\cdot(ax+b)\) incorrectly when \(b\ne0\): the term \(\frac{ax+b}{a}\arctan(ax+b)\) is \(\bigl(x+\frac ba\bigr)\arctan(ax+b)\), not \(x\arctan(ax+b)\).

A third error concerns the quadratic: \(1+(ax+b)^2\ne1+a^2x^2+b^2\). The cross term \(2abx\) must be kept when the quadratic is expanded.

Quick Reference

Interpreting the Result

The affine argument stretches and shifts the graph of \(\arctan\). A slope \(a>1\) compresses the curve horizontally, so the integrand climbs to its limiting height \(\frac\pi2\) more quickly, and the area under it over a fixed interval increases. A shift \(b\) slides the curve left or right. Because the area scales like \(\frac1a\) under a horizontal stretch, the factor \(\frac1a\) in the antiderivative is the same bookkeeping that appears in any change of variable.

Practice Problems with Answers

Try each problem before reading the answer. Every answer below was checked by differentiation or by numerical integration.

1. Find an antiderivative of \(\arctan(3x)\). Hint: take \(a=3\) and multiply the bracket by \(\frac13\). Answer: \(x\arctan(3x)-\frac16\ln(1+9x^2)\)
2. Evaluate \(\int_0^1\arctan(2x)\,dx\). Hint: use the antiderivative from the second worked example. Answer: \(\arctan2-\frac14\ln5\approx0.705\)
3. Find an antiderivative of \(\arctan(x-2)\). Hint: expand \(1+(x-2)^2\). Answer: \((x-2)\arctan(x-2)-\frac12\ln(x^2-4x+5)\)
4. Evaluate \(\int_0^1\arctan(3x-1)\,dx\). Hint: here \(a=3\), \(b=-1\), so the bracket is evaluated at \(2\) and \(-1\). Answer: \(\frac13\bigl[2\arctan2-\frac\pi4-\frac12\ln5+\frac12\ln2\bigr]\)

Two Routes to the Same Answer

Instead of substituting first, apply integration by parts directly with \(u=\arctan(ax+b)\) and \(dv=dx\). Then \(du=\frac{a\,dx}{1+(ax+b)^2}\) and \(v=x\), so the formula produces \(x\arctan(ax+b)-\int\frac{ax}{1+(ax+b)^2}\,dx\). The remaining integral is a rational function whose numerator is not exactly the derivative of the denominator when \(b\ne0\). One must write \(ax=(ax+b)-b\), integrate the logarithmic part, and then handle the leftover \(-b\) term as an inverse tangent. That produces the same answer, but the substitution route is shorter and less error prone.

The comparison is instructive. When \(b=0\), both routes are equally short. When \(b\ne0\), the substitution route hides the extra arctangent term inside the combination \((ax+b)\arctan(ax+b)\), which is why it is preferred for affine arguments.

Domain and Behavior

The arctangent is defined for every real number, so \(\arctan(ax+b)\) is continuous everywhere and every definite integral exists. There is no restriction on the interval, unlike the logarithm and inverse sine integrals, where the argument must stay in a restricted range. The logarithm \(\ln(1+(ax+b)^2)\) in the answer is also defined for every real \(x\), because its argument is at least \(1\). This is why no absolute value signs or interval conditions appear in the result.

For large \(|x|\) the integrand approaches \(\pm\frac\pi2\), so the antiderivative grows roughly like \(\frac{\pi}{2}|x|\). The logarithm contributes only a slow correction of order \(\frac{1}{a}\ln|ax|\).

A Sanity Check You Can Always Run

Before trusting any answer of this type, test it at a convenient point. For \(\arctan(x+1)\) at \(x=0\), the integrand equals \(\arctan1=\frac\pi4\approx0.785\). Differentiating the claimed antiderivative at \(x=0\) must give the same number. The product-rule term contributes \(\arctan1+\frac{1}{2}\), the logarithm contributes \(-\frac12\), and the sum is \(\frac\pi4\). A mismatch at a single point immediately reveals a missing factor or a sign error, which is far faster than re-deriving the whole calculation.

Where This Fits in the Course

This tutorial belongs to the module “Integration by Parts I”. The tutorial just before it is “Integrating The Inverse Sine And Cosine,” and the one that follows is “Integrating x Times The Inverse Tangent.” It generalizes the arctangent formula just derived and prepares for products such as \(x\arctan x\).