Let's do some integration!

Calculus Level 4

∫ 0 1 ln ⁡ ( x + 1 ) x 2 + 1 d x = π a ln ⁡ b \large \int_0^1 \frac {\ln(x+1)}{x^2+1}\ dx = \frac \pi a \ln b

The equation above holds true for some real numbers a a and b b . Determine the value of a + b . a+b.


The answer is 10.

Juvin Abayon by Juvin Abayon , 24, Philippines

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2 solutions

Harsh Shrivastava
Jul 23, 2017

Trigonometric substitution is good but very lengthy. A quick alternative would be to substitute x = 1 − y 1 + y x = \frac{1-y}{1+y} then the integral becomes very easy to solve.

The motivation behind the substitution is that by doing this substitution, we can ,by using properties of log, get the original integrand back.

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Is it lengthy. I = ∫ l n ( t a n ( θ ) + 1 ) d θ = ∫ l n ( t a n ( π / 4 − θ ) + 1 ) = π / 4 l n 2 − ∫ l n ( 1 + t a n θ ) I=\int ln(tan(\theta )+1)d\theta =\int ln(tan(π/4-\theta )+1)=π/4ln2-\int ln(1+tan\theta ) .So 2 I = π / 4 l n 2 2I=π/4ln2

Spandan Senapati - 3 years, 10 months ago

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No, I was referring to Juvin's solution. And yeah your approach is much better and I also wanted to show a new substitution method which is useful in some other types of problems.

Harsh Shrivastava - 3 years, 10 months ago
Juvin Abayon
Jul 23, 2017

Having a change in variables, let x = t a n θ x = tanθ .

Then, x 2 + 1 = s e c 2 θ x^2+1 = sec^2θ and d x = s e c 2 d θ dx = sec^2 dθ as well. In addition, a change in limits would give the new upper and lower limits as a r c t a n 1 = π / 4 arctan 1 = π/4 and a r c t a n 0 = 0 arctan 0 = 0 , respectively. Denoting the value of the integral as I, the solution are as shown in the photo. The determination of a and b yields 8 8 and 2 2 , respectively, and at once, a + b = 10 a+b = 10

PS. The question, and the corresponding solution, were all done in Microsoft Word. I do apologize for being unable to do the entire solution in LaTeX.

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Similar solution but slightly simpler.

I = ∫ 0 1 ln ⁡ ( x + 1 ) x 2 + 1 d x Let x = tan ⁡ θ ⟹ d x = ( tan ⁡ 2 θ + 1 ) d θ = ∫ 0 π 4 ln ⁡ ( tan ⁡ θ + 1 ) d θ Using ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x = 1 2 ∫ 0 π 4 ln ⁡ ( tan ⁡ θ + 1 ) + ln ⁡ ( tan ⁡ ( π − θ ) + 1 ) d θ = 1 2 ∫ 0 π 4 ln ⁡ ( tan ⁡ θ + 1 ) + ln ⁡ ( 1 − tan ⁡ θ tan ⁡ θ + 1 + 1 ) d θ = 1 2 ∫ 0 π 4 ln ⁡ ( tan ⁡ θ + 1 ) + ln ⁡ ( 2 tan ⁡ θ + 1 ) d θ = 1 2 ∫ 0 π 4 ln ⁡ ( tan ⁡ θ + 1 ) + ln ⁡ 2 − ln ⁡ ( tan ⁡ θ + 1 ) d θ = 1 2 ∫ 0 π 4 ln ⁡ 2 d θ = π 8 ln ⁡ 2 \begin{aligned} I & = \int_0^1 \frac {\ln (x+1)}{x^2+1} \ dx & \small \color{#3D99F6} \text{Let }x = \tan \theta \implies dx = (\tan^2 \theta +1) \ d\theta \\ & = \int_0^\frac \pi 4 \ln (\tan \theta +1) \ d\theta & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\frac \pi 4 \ln (\tan \theta +1) + \ln (\tan (\pi - \theta) +1) \ d\theta \\ & = \frac 12 \int_0^\frac \pi 4 \ln (\tan \theta +1) + \ln \left(\frac {1 - \tan \theta}{\tan \theta + 1} +1 \right) \ d\theta \\ & = \frac 12 \int_0^\frac \pi 4 \ln (\tan \theta +1) + \ln \left(\frac 2{\tan \theta + 1} \right) \ d\theta \\ & = \frac 12 \int_0^\frac \pi 4 \ln (\tan \theta +1) + \ln 2 - \ln (\tan \theta + 1) \ d\theta \\ & = \frac 12 \int_0^\frac \pi 4 \ln 2 \ d\theta \\ & = \frac \pi 8 \ln 2 \end{aligned}

⟹ a + b = 8 + 2 = 10 \implies a+b = 8+2=\boxed{10} .

Chew-Seong Cheong - 3 years, 10 months ago

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