Math Archive (8)

Calculus Level 4

∫ 1 + x 2 3 − 4 2 x + 4 x 4 d x = a x + b + c ( tan ⁡ − 1 ( x + d ) ) + c o n s t a n t \large \int\frac{1+x^{2}}{3-4\sqrt{2}x+4x^{4}} dx =\frac{a}{x+b}+c \left(\tan^{-1}(x+d)\right)+\color{grey}constant Find a + b + c + d a+b+c+d .

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The answer is 0.

Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

I = ∫ x 2 + 1 4 x 4 − 4 2 x + 3 d x = ∫ x 2 + 1 ( 2 x 2 − 2 2 x + 3 ) ( 2 x 2 + 2 2 x + 1 ) d x By partial fraction decomposition = 1 4 ∫ ( 1 2 x 2 − 2 2 x + 1 + 1 2 x 2 + 2 2 x + 3 ) d x = 1 4 ∫ ( 1 ( 2 x − 1 ) 2 + 1 ( 2 x + 1 ) 2 + 2 ) d x = 1 8 ∫ ( 1 ( x − 1 2 ) 2 + 1 ( x + 1 2 ) 2 + 1 ) d x = − 1 8 x − 1 2 + 1 8 tan ⁡ − 1 ( x + 1 2 ) + C where C is the constant of integration. \begin{aligned} I & = \int \frac {x^2+1}{4x^4-4\sqrt 2 x+3}dx \\ & = \int \frac {x^2+1}{\left(2x^2-2\sqrt 2 x+3\right)\left(2x^2+2\sqrt 2 x+1\right)}dx & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = \frac 14 \int \left(\frac 1{2x^2-2\sqrt 2 x+1} + \frac 1{2x^2+2\sqrt 2 x+3} \right) dx \\ & = \frac 14 \int \left(\frac 1{(\sqrt 2x-1)^2} + \frac 1{(\sqrt 2x+1)^2+2} \right) dx \\ & = \frac 18 \int \left(\frac 1{(x-\frac 1{\sqrt 2})^2} + \frac 1{(x+ \frac 1{\sqrt 2})^2+1} \right) dx \\ & = \frac {-\frac 18}{x- \frac 1{\sqrt 2}} + \frac 18 \tan^{-1} \left(x + \frac 1{\sqrt 2} \right) + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \end{aligned}

Therefore, a + b + c + d = − 1 8 − 1 2 + 1 8 + 1 2 = 0 a+b+c+d = - \dfrac 18 - \dfrac 1{\sqrt 2} + \dfrac 18 + \dfrac 1{\sqrt 2} = \boxed 0 ,

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