Solve-dx-1-x-2-3-

Question Number 132082 by liberty last updated on 11/Feb/21

Solve \int \frac{dx}{{(1 + x^{2})}^{3}} ?

Answered by EDWIN88 last updated on 11/Feb/21

Ostrogradski method

Consider \frac{d}{dx} (\frac{{ax}^{3} + bx}{{(x^{2} + 1)}^{2}}) = \frac{- {ax}^{4} + (3 a - 3 b) x^{2} + b}{{(x^{2} + 1)}^{3}}

\int \frac{d}{dx} [\frac{{ax}^{3} + bx}{{(x^{2} + 1)}^{2}}] = \int \frac{- {ax}^{4} + (3 a - 3 b) x^{2} + b}{{(x^{2} + 1)}^{3}} dx + \int \frac{c}{x^{2} + 1} dx

comparing coefficients

1 = - {ax}^{4} + (3 a - 3 b) x^{2} + c {(x^{2} + 1)}^{2} + b

1 = (c - a) x^{4} + (3 a - 3 b + 2 c) x^{2} + b + c

we get a = c, b + c = 1, 3 a - 3 b + 2 c = 0

then a = c = \frac{3}{8}; b = \frac{5}{8}

therefore I = \frac{1}{8} [\frac{{3 x}^{3} + 5 x}{{(x^{2} + 1)}^{2}}] + \frac{3}{8} \int \frac{dx}{x^{2} + 1}

I = \frac{1}{8} [\frac{{3 x}^{3} + 5 x}{{(x^{2} + 1)}^{2}}] + \frac{3}{8} \arctan (x) + c

please check

Answered by liberty last updated on 11/Feb/21

by Ostrogradski method

let \int \frac{dx}{{(x^{2} + 1)}^{3}} = \frac{{ax}^{3} + bx}{{(x^{2} + 1)}^{2}} + \int \frac{c}{x^{2} + 1} dx

differentiating both sides

\frac{1}{{(x^{2} + 1)}^{3}} = \frac{({3 ax}^{2} + b) {(x^{2} + 1)}^{2} - 4 x (x^{2} + 1) ({ax}^{3} + bx)}{{(x^{2} + 1)}^{4}} + \frac{c}{x^{2} + 1}

\frac{1}{{(x^{2} + 1)}^{3}} = \frac{({3 ax}^{2} + b) (x^{2} + 1) - ({4 ax}^{4} + {4 bx}^{2})}{{(x^{2} + 1)}^{3}} + \frac{c {(x^{2} + 1)}^{2}}{{(x^{2} + 1)}^{3}}

1 = {3 ax}^{4} + {3 ax}^{2} + {bx}^{2} + b - {4 ax}^{4} - {4 bx}^{2} + {cx}^{4} + {2 cx}^{2} + c

1 = (- a + c) x^{4} + (3 a - 3 b + 2 c) x^{2} + b + c

we get c = a; b + c = 1; 3 a - 3 b + 2 c = 0

\Rightarrow 5 c - 3 (1 - c) = 0; 8 c = 3

c = a = \frac{3}{8} and b = \frac{5}{8}

then the integral become

\int \frac{dx}{{(x^{2} + 1)}^{3}} = \frac{{3 x}^{3} + 5 x}{8 {(x^{2} + 1)}^{2}} + \int \frac{3}{8} \frac{dx}{(x^{2} + 1)}

= \frac{{3 x}^{3} + 5 x}{8 {(x^{2} + 1)}^{2}} + \frac{3}{8} \arctan x + c

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