Evaluate ∫₀^1 tan^-1 x dx.

Definite Integrals & Area · Question 33 of 71

Question 33

HardOriginal question

Evaluate $\int_0^1 \tan^{-1} x\,dx$ .

- \log 2 + \frac{\pi}{4}

Correct answerYour answer

- \frac{\log 2}{2} + \frac{\pi}{4}

Correct answerYour answer

\frac{\pi}{4}

Correct answerYour answer

\frac{\log 2}{2} + \frac{\pi}{4}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

By parts with $u = \tan^{-1} x$ , $dv = dx$ : $\int \tan^{-1} x\,dx = x\tan^{-1} x - \frac{1}{2}\log(1 + x^2)$ .
From 0 to 1: $\frac{\pi}{4} - \frac{1}{2}\log 2$ .

Worth noting

$\frac{\pi}{4}$ stops at $uv$ and leaves out $-\int v\,du$ .

Ten questions from Definite Integrals & Area, at your own pace.

Ten questions from Definite Integrals & Area, at your own pace.