Evaluate: integral from 0 to 1 of $x \tan^{-1} x \, dx$ .

Q: Evaluate: integral from 0 to 1 of $x \tan^{-1} x \, dx$.

The value of the definite integral is **$(\pi - 2)/4$**. **Method:** Using integration by parts with $u = \tan^{-1}x$ and $dv = x \, dx$ gives $(x^2/2)\tan^{-1}x - \tfrac{1}{2}\int x^2/(1+x^2) \, dx$. Simplifying the remaining integral using $x^2/(1+x^2) = 1 - 1/(1+x^2)$ and evaluating from 0 to 1 gives $\pi/4 - 1/2$, i.e. **$(\pi - 2)/4$**.

Q: What is the final numeric value of the integral?

It equals $(\pi - 2)/4$, approximately 0.2854.

Integrals

Definite Integral by Parts

Official Answer

The value of the definite integral is $(\pi - 2)/4$ .

Method: Using integration by parts with $u = \tan^{-1}x$ and $dv = x \, dx$ gives $(x^2/2)\tan^{-1}x - \tfrac{1}{2}\int x^2/(1+x^2) \, dx$ . Simplifying the remaining integral using $x^2/(1+x^2) = 1 - 1/(1+x^2)$ and evaluating from 0 to 1 gives $\pi/4 - 1/2$ , i.e. $(\pi - 2)/4$ .

integration by partsdefinite integraltan inverse xILATE rulelimits of integrationalgebraic and inverse trig function

Marking Scheme

11 mark: correct choice of parts ( $u = \tan^{-1}x$ , $dv = x \, dx$ ) and correct du, v.
21 mark: correct simplification of $x^2/(1+x^2) = 1 - 1/(1+x^2)$ and setting up the remaining integral.
31 mark: correct evaluation of limits and final answer $(\pi - 2)/4$ .

Hint

Take $u = \tan^{-1}x$ , $dv = x \, dx$ ; use integration by parts, then simplify $x^2/(1+x^2) = 1 - 1/(1+x^2)$ .

Quick Oral Answer

Using integration by parts with $u = \tan^{-1}x$ and $v = x^2/2$ , the integral evaluates to $(\pi - 2)/4$ after simplifying $x^2/(1+x^2)$ .

Analysis & Explanation

This is a classic ILATE-rule integration-by-parts question combined with definite-integral evaluation.

Concept: Since $\tan^{-1}x$ is inverse trigonometric and $x$ is algebraic, ILATE priority makes $\tan^{-1}x$ the 'first function' ( $u$ ) and $x \, dx$ the 'second function' ( $dv$ ) — this ordering keeps the resulting integral simple.

Exam trap: A common error is choosing u = x instead of u = tan⁻¹x, or forgetting to simplify $x^2/(1+x^2)$ before integrating, leaving an unresolved rational integral.

Real-world link: Definite integrals of this form appear in computing weighted-average phase angles and moment calculations in engineering problems involving arctangent-weighted distributions.

Common Mistakes

1Choosing x as the first function (u) instead of $\tan^{-1}x$ , violating the ILATE priority and complicating the remaining integral.
2Forgetting to rewrite $x^2/(1+x^2)$ as $1 - 1/(1+x^2)$ before integrating, leaving the integral unresolved.
3Sign errors while substituting the limits 0 and 1 into the $\tan^{-1}x$ and x terms.

Interesting Facts

Integration by parts is derived directly from the product rule of differentiation, reversed and rearranged.

Definite integrals involving tan⁻¹x often produce results containing π, linking algebraic areas to circular geometry.

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Frequently Asked Questions

Why is $\tan^{-1}x$ chosen as the first function in integration by parts?

By the ILATE priority rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential), inverse trigonometric functions are taken as u before algebraic functions, since their derivative is algebraic and simplifies the resulting integral.

What is the final numeric value of the integral?

It equals $(\pi - 2)/4$ , approximately 0.2854.