An Epic Integral

Calculus Level 2

If ∫ 1 x ( x + 1 ) d x = A ln ⁡ ∣ x ∣ + B ln ⁡ ∣ x + 1 ∣ + C \displaystyle \int \frac{1}{x(x+1)} \ dx = A\ln|x|+B\ln|x+1| + C , find A − B A-B .


The answer is 2.

Solden Stoll by Solden Stoll , 16, USA

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1 solution

Chew-Seong Cheong
Mar 25, 2019

I = ∫ 1 x ( x + 1 ) d x = ∫ ( 1 x − 1 x + 1 ) d x = ln ⁡ ∣ x ∣ − ln ⁡ ∣ x + 1 ∣ + C where C is the constant of integration. \begin{aligned} I & = \int \frac 1{x(x+1)}\ dx \\ & = \int \left( \frac 1x - \frac 1{x+1}\right) dx \\ & = \ln |x| - \ln |x+1| + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \end{aligned}

Therefore, A − B = 1 − ( − 1 ) = 2 A-B = 1-(-1) = \boxed 2 .

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