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Differentiability

√4× 4+3 × 2+2...

Question

$\int \frac{4 x^{4} + 3}{(x^{2} + 2) (x^{2} + 3) (x^{2} + 4)} d x$

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Solution

$We have, I = \int \frac{(4 x^{4} + 3) d x}{(x^{2} + 2) (x^{2} + 3) (x^{2} + 4)} Putting x^{2} = t Then, \frac{4 x^{4} + 3}{(x^{2} + 2) (x^{2} + 3) (x^{2} + 4)} = \frac{4 t^{2} + 3}{(t + 2) (t + 3) (t + 4)} Let \frac{4 t^{2} + 3}{(t + 2) (t + 3) (t + 4)} = \frac{A}{t + 2} + \frac{B}{t + 3} + \frac{C}{t + 4} \Rightarrow \frac{4 t^{2} + 3}{(t + 2) (t + 3) (t + 4)} = \frac{A (t + 3) (t + 4) + B (t + 2) (t + 4) + C (t + 2) (t + 3)}{(t + 2) (t + 3) (t + 4)} \Rightarrow 4 t^{2} + 3 = A (t + 3) (t + 4) + B (t + 2) (t + 4) + C (t + 2) (t + 3) Putting t + 3 = 0 \Rightarrow t = - 3 ∴ 4 \times {(- 3)}^{2} + 3 = B (- 3 + 2) (- 3 + 4) \Rightarrow 39 = B (- 1) \Rightarrow B = - 39 Putting t + 2 = 0 \Rightarrow t = - 2 ∴ 4 {(- 2)}^{2} + 3 = A (- 2 + 3) (- 2 + 4) \Rightarrow 19 = A \times 1 \times 2 \Rightarrow A = \frac{19}{2} Let t + 4 = 0 \Rightarrow t = - 4 ∴ 4 \times {(- 4)}^{2} + 3 = C (- 4 + 2) (- 4 + 3) \Rightarrow 67 = C (- 2) (- 1) \Rightarrow C = \frac{67}{2} ∴ \frac{4 t^{2} + 3}{(t + 2) (t + 3) (t + 4)} = \frac{19}{2 (t + 2)} - \frac{39}{t + 3} + \frac{67}{2 (t + 4)} \Rightarrow \frac{4 x^{4} + 3}{(x^{2} + 2) (x^{2} + 3) (x^{2} + 4)} = \frac{19}{2 (x^{2} + 2)} - \frac{39}{x^{2} + 3} + \frac{67}{2 (x^{2} + 4)} ∴ I = \frac{19}{2} \int \frac{d x}{x^{2} + {(\sqrt{2})}^{2}} - 39 \int \frac{d x}{x^{2} + {(\sqrt{3})}^{2}} - \frac{67}{2} \int \frac{d x}{x^{2} + 2^{2}} = \frac{19}{2} \times \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{x}{\sqrt{2}}) - \frac{39}{\sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}}) - \frac{67}{2} \times \frac{1}{2} \tan^{- 1} (\frac{x}{2}) + C = \frac{19}{2 \sqrt{2}} \tan^{- 1} (\frac{x}{\sqrt{2}}) - \frac{39}{\sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}}) - \frac{67}{4} \tan^{- 1} (\frac{x}{2}) + C$

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Similar questions

\int \frac{1}{(2 x^{2} + 3) \sqrt{x^{2} - 4}} d x

\int \frac{x^{2} (x^{4} + 4)}{x^{2} + 4} d x

Q. Observe the following pattern

(1 \times 2) + (2 \times 3) = \frac{2 \times 3 \times 4}{3}

(1 \times 2) + (2 \times 3) + (3 \times 4) = \frac{3 \times 4 \times 5}{3}

(1 \times 2) + (2 \times 3) + (3 \times 4) + (4 \times 5) = \frac{4 \times 5 \times 6}{3}

and find the value of
(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + (5 × 6)

(1) 4² + 2⁴− 3²

(2) 3³ − 2²+ 2⁰

(3) 5³ − 2² + 4⁰

(4) 6¹ + 5⁰ − 4¹

(5) (−2)² + 1² − 2²

I = \int (\frac{1}{x^{1 / 2} - x^{1 / 4}} + \frac{log (1 + x^{1 / 4})}{x^{1 / 2} + x^{1 / 4}}) d x

, then I equals

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