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Question Number 132333 by liberty last updated on 13/Feb/21

very nice integral

\int \frac{{4 x}^{3} + {4 x}^{2} + 4 x + 3}{(x^{2} + 1) {(x^{2} + x + 1)}^{2}} dx ?

Answered by EDWIN88 last updated on 13/Feb/21

well \dots let me try

Ostrogradsky method

the integral has form

\int \frac{{4 x}^{3} + {4 x}^{2} + 4 x + 3}{(x^{2} + 1) {(x^{2} + x + 1)}^{2}} dx = \frac{ax + b}{x^{2} + x + 1} + \int \frac{{cx}^{3} + {dx}^{2} + ex + f}{(x^{2} + 1) (x^{2} + x + 1)} dx

differentiating both sides

\frac{{4 x}^{3} + {4 x}^{2} + 4 x + 3}{(x^{2} + 1) {(x^{2} + x + 1)}^{2}} = \frac{a (x^{2} + x + 1) - (ax + b) (2 x + 1)}{{(x^{2} + x + 1)}^{2}} + \frac{{cx}^{3} + {dx}^{2} + ex + f}{(x^{2} + 1) (x^{2} + x + 1)}

after solving the coefficients we get

a = 1, b = - 1, c = 0, d = 1, e = 1, f = 1

then I = \frac{x - 1}{x^{2} + x + 1} + \int \frac{x^{2} + x + 1}{(x^{2} + 1) (x^{2} + x + 1)} dx

I = \frac{x - 1}{x^{2} + x + 1} + \int \frac{dx}{x^{2} + 1} = \frac{x - 1}{x^{2} + x + 1} + \arctan (x) + C

Commented by liberty last updated on 13/Feb/21

sch \ddot{o n e} L \ddot{o s u n g}

Commented by SLVR last updated on 13/Feb/21

g r e a t j o b \dots n i c e s i r

Commented by SLVR last updated on 13/Feb/21

b u t o n d i f f e r e n t i a i o n o f R H S 1 s t p a r t d e n o m i n a t o r

h a s n o s q a r e \dots h o w \dots t o p r o c e e d . .

Commented by EDWIN88 last updated on 13/Feb/21

oh it is only typo

Commented by SLVR last updated on 13/Feb/21

w e l c o m e s i r . .

Answered by EDWIN88 last updated on 13/Feb/21

\int \frac{P (x)}{Q (x)} dx = \frac{P_{1} (x)}{Q_{1} (x)} + \int \frac{P_{2} (x)}{Q_{2} (x)} dx

where Q_{1} (x) is \gcd of Q (x) and Q^{'} (x)

Q_{2} (x) = \frac{Q (x)}{Q_{1} (x)} and P_{1} (x) and P_{2} (x) are

unknown polynomials of degree one less

than Q_{1} (x) and Q_{2} (x) respectively .

from the equation Q (x) = (x^{2} + 1) {(x^{2} + x + 1)}^{2}

Q^{'} (x) = 2 x {(x^{2} + x + 1)}^{2} + 2 (x^{2} + 1) (2 x + 1) (x^{2} + x + 1)

the \gcd of Q (x) and Q^{'} (x) is Q_{1} (x) = x^{2} + x + 1

and Q_{2} (x) = \frac{Q (x)}{Q_{1} (x)} = (x^{2} + 1) (x^{2} + x + 1)

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