Integration Arctan x

Calculus Level 2

∫ arctan ⁡ x d x \large \int \arctan x\, dx

Evaluate the indefinite integral above.

Notation : C C denotes the arbitrary constant of integration.

arctan ⁡ x − 1 2 ln ⁡ ( x 2 + 1 ) \arctan x-\frac{1}{2}\ln(x^2+1) x arctan ⁡ x − 1 2 ln ⁡ ( x 2 + 1 ) + C x \arctan x-\frac{1}{2}\ln(x^2+1) + C x arctan ⁡ x + 1 2 ln ⁡ ( x 2 + 1 ) + C x \arctan x+\frac{1}{2}\ln(x^2+1) + C x arctan ⁡ x − 1 3 ln ⁡ ( x + 1 ) + C x \arctan x-\frac{1}{3}\ln(x+1) + C

Dimitris Kouroupos by dimitris kouroupos , 23, Greece

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2 solutions

Chew-Seong Cheong
Apr 27, 2017

I = ∫ arctan ⁡ x d x Let θ = arctan ⁡ x , tan ⁡ θ = x , sec ⁡ 2 θ d θ = d x = ∫ θ sec ⁡ 2 θ d θ By integration by parts = θ tan ⁡ θ − ∫ tan ⁡ θ d θ = θ tan ⁡ θ − ∫ sin ⁡ θ cos ⁡ θ d θ Note that d cos ⁡ θ = − sin ⁡ θ d θ = θ tan ⁡ θ + ∫ 1 cos ⁡ θ d cos ⁡ θ = θ tan ⁡ θ + ln ⁡ ( cos ⁡ θ ) + C where C is the constant of integration. = x arctan ⁡ x + ln ⁡ ( 1 x 2 + 1 ) + C Note that tan ⁡ θ = x ⟹ cos ⁡ θ = 1 x 2 + 1 = x arctan ⁡ x − 1 2 ln ⁡ ( x 2 + 1 ) + C \begin{aligned} I & = \int \arctan x \ dx & \small \color{#3D99F6} \text{Let }\theta = \arctan x, \ \tan \theta = x, \ \sec^2 \theta \ d\theta = dx \\ & = \int \theta \sec^2 \theta \ d\theta & \small \color{#3D99F6} \text{By integration by parts} \\ & = \theta \tan \theta - \int \tan \theta \ d\theta \\ & = \theta \tan \theta \ {\color{#3D99F6}-} \int \frac {\color{#3D99F6}\sin \theta}{\cos \theta} \color{#3D99F6}d\theta & \small \color{#3D99F6} \text{Note that } d \cos \theta = - \sin \theta \ d \theta \\ & = \theta \tan \theta \ {\color{#3D99F6}+} \int \frac 1{\cos \theta} \color{#3D99F6}d \cos \theta \\ & = \theta \tan \theta + \ln (\cos \theta) + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ & = x \arctan x + \ln \left(\sqrt{\frac 1{x^2+1}}\right) + C & \small \color{#3D99F6} \text{Note that }\tan \theta = x \implies \cos \theta = \sqrt{\frac 1{x^2+1}} \\ & = \boxed{x \arctan x - \dfrac 12 \ln \left(x^2+1\right) + C} \end{aligned}

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With the help of the factorial method we are catching up in the final answer

∫ A r c t a n x d x . = ∫ ( x ) ′ A r c t a n x d x . = x A r c t a n x − ∫ x x 2 + 1 d x . = x A r c t a n x − 1 2 l n ( x 2 + 1 ) + C \int Arc tan x\, dx . = \int (x)' Arc tan x\, dx . =x Arctan x- \int \frac{x}{x^2+1}dx.=x Arc tan x-\frac{1}{2}ln(x^2+1) + C

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