'A' comes first

Calculus Level 2

If ∫ 1 e − 2 x + e 2 x d x = A 6 t a n − 1 ( e 2 x ) + C \displaystyle\int\dfrac{1}{e^{-2x}+e^{2x}}dx=\dfrac{A}{6}tan^{-1}(e^{2x})+C ,then what is the value of A ? A?


The answer is 3.

Skanda Prasad by Skanda Prasad , 20, India

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1 solution

Tapas Mazumdar
Oct 18, 2017

I = ∫ 1 e − 2 x + e 2 x d x = 1 2 ⋅ ∫ 2 e 2 x ( e 2 x ) 2 + 1 d x Substitute e 2 x = u ⟹ 2 e 2 x d x = d u = 1 2 ⋅ arctan ⁡ ( e 2 x ) + C \begin{aligned} \displaystyle I &= \int \dfrac{1}{e^{-2x} + e^{2x}} \,dx \\ \displaystyle &= \dfrac 12 \cdot \int \dfrac{2e^{2x}}{ {\left( e^{2x} \right)}^2 +1} \,dx & \small\color{#3D99F6}{\text{Substitute } e^{2x} = u \implies 2e^{2x} \,dx = \,du} \\ \displaystyle &= \dfrac 12 \cdot \arctan \left( e^{2x} \right) + C \end{aligned}

∴ 1 2 = 3 6 ⟹ A = 3 \therefore \dfrac 12 = \dfrac 36 \implies \boxed{A=3}

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