Question

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Solve this integration

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Answer 1

HEY BUDDY

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HERE IS YOUR SOLUTION

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COMMENT BELOW⬇⬇⬇⬇ FOR DOUBTS

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Answer 2

HELLO DEAR,

$\bold{⇒\int{\frac{(x^2 + 1)(x^2 + 2)}{(x^2 + 3)(x^2 + 4)}}\,dx}$

$\bold{⇒\int{\frac{[(x^2 + 3) - 2][(x^2 + 4) - 2]}{(x^2 + 3)(x^4)}}\,dx}$

$\bold{⇒\int{\frac{(x^2 +3)(x^2+4) - 2(x^2+3) - 2(x^2+4) + 4}{(x^2 + 3)(x^2+4)}}\,dx}$

$\bold{⇒\int{[1 - \frac{2(x^2+3)}{(x^2+3)(x^2+4)} - \frac{2(x^2+4)}{(x^2+3)(x^2+4)} + \frac{4}{(x^2+3)(x^2+4)}]}\,dx}$

$\bold{⇒\int{[1 - \frac{2}{x^2+4} - \frac{2}{x^2+4} + \frac{4}{(x^2+3)(x^2+4)}]}\,dx}$

$\bold{⇒\int{[1 - \frac{2}{x^2+4} - \frac{2}{x^2+3} + 4\frac{(x^2 + 4) - (x^2 + 3)}{(x^2+ 3)(x^2+4)}]}\,dx}$

$\bold{⇒\int{[1 - 2\frac{1}{x^2 + 2^2} - 2\frac{1}{x^2 + (\sqrt{3})^2} + 4\frac{1}{x^2 + 3} - 4\frac{1}{x^2+4}]}\,dx}$

$\bold{⇒\int{[1 - 6\frac{1}{x^2 + 2^2} + 2\frac{1}{x^2 + (\sqrt{3})^2}]}\,dx}$

$\bold{⇒\int{1}\,dx - 6\int{\frac{1}{(x^2 + 2^2)}}\,dx + 2\int{\frac{1}{x^2 + (\sqrt{3})^2}}\,dx}$

$\bold{⇒x - \frac{6}{2}tan^{-1}\frac{x}{2} + \frac{2}{\sqrt{3}}tan^{-1}\frac{x}{\sqrt{3}} + c}$

I HOPE ITS HELP YOU DEAR,
THANKS