--- title: "Show ∫ log(1+x)/(1+x²) dx = (π/8) log 2" subject: "Mathematics I" chapter: 2 key: "semester_1/mathematics_i:integral-log-1-plus-x-over-1-plus-x2" source: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-log-1-plus-x-over-1-plus-x2 site: BCT Engineering --- # Show ∫ log(1+x)/(1+x²) dx = (π/8) log 2 Mathematics I · Chapter 2: Integration and its Applications - Asked in: 2070 Asar - Video: [Show that Integral 0 to 1 log(1+x)/(1+x^2) dx = π/8 log2 | Integration of log(1+x)/(1+x^2) dx](https://www.youtube.com/watch?v=j4NHFzmbVW8) — Nitin Kumar Gupta Classes - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-log-1-plus-x-over-1-plus-x2 - Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2 ## Question Show that ∫₀¹ log(1+x)/(1+x²) dx = (π/8) log 2. ## Short answer Substitute x = tan θ to get I = ∫_0^{π/4} log(1+tan θ)dθ. Replace θ by π/4–θ to obtain I = (π/4)log 2 – I, giving 2I = (π/4)log 2, so I = (π/8)log 2. ## Answer Let I = ∫₀¹ log(1 + x)/(1 + x²) dx. Put x = tan θ, dx = sec²θ dθ, 1 + x² = sec²θ. Limits: 0 to π/4. I = ∫₀^(π/4) log(1 + tan θ) dθ Use ∫₀^a f(θ)dθ = ∫₀^a f(a − θ)dθ with a = π/4. Since tan(π/4 − θ) = (1 − tan θ)/(1 + tan θ): 1 + tan(π/4 − θ) = 2/(1 + tan θ) ```text I = ∫₀^(π/4) log[2/(1 + tan θ)] dθ = ∫₀^(π/4) log 2 dθ − ∫₀^(π/4) log(1 + tan θ) dθ = (π/4) log 2 − I ``` So 2I = (π/4) log 2. **I = (π/8) log 2.** Shown. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. x = tan θ reduces the integral to A. integral log(1+sin θ) dθ from 0 to π/2 B. integral log(1 + tan θ) dθ from 0 to π/4 C. integral log(sec θ) dθ D. integral log(tan θ) dθ Answer: B. dx/(1+x^2) = dθ. (Key point: Substitution) 2. 1 + tan(π/4 - θ) equals A. (1 - tan θ) B. 1/(1+tan θ) C. 2/(1 + tan θ) D. 2 tan θ Answer: C. tan(π/4 - θ) = (1 - tan θ)/(1 + tan θ). (Key point: Identity) 3. This leads to I = (π/4) log 2 - I because A. the integral of log 2 vanishes B. I is odd C. the limits cancel D. log of the quotient splits into log 2 minus the original integrand Answer: D. The second term is -I. (Key point: Equation) 4. I equals A. (π/8) log 2 B. (π/4) log 2 C. (π/2) log 2 D. π log 2 Answer: A. 2I = (π/4) log 2. (Key point: Result) ## Related in Mathematics I - [DUIS: ∫ log((a+b sin x)/(a−b sin x)) dx/sin x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aduis-log-a-plus-b-sinx) — Mathematics I, chapter 2, asked 9× - [DUIS: ∫ log(1+a²x²)/(1+b²x²) dx](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aduis-log-1-plus-a2x2) — Mathematics I, chapter 2, asked 4× - [Solve x dy/dx + y log y = xy eˣ](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-x-dy-dx-y-log-y) — Mathematics I, chapter 4, asked 4× - [Solve x log x dy/dx + y = 2 log x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-x-log-x-dy-dx) — Mathematics I, chapter 4, asked 4× - [Show ∫ log(1+x²)/(1+x²) dx = π log 2](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aintegral-log-1-plus-x2) — Mathematics I, chapter 2, asked 2× - [Show ∫ x log(sin x) dx](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aintegral-x-log-sin-x) — Mathematics I, chapter 2 ## Related in other subjects - [Gauss 2-point and 3-point: log(x+1)+sin 2x](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Agauss-2pt-3pt-log-sin) — Numerical Method, chapter 5 - [Triple integral of log z](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Atriple-integral-log-z) — Mathematics, chapter 2