Question

integration of x square + 1 into X square + 2 divide by X square + 3 into X square + 4

Answer 1

Answer:

x + 2tan^-1(x/√3)/√3 – 6tan^-1(x/2)/2 + C

Step-by-step explanation:

Integrate (x²+1)(x²+2) / (x²+3)(x²+4)

F(x) = (x²+1)(x²+2) / (x²+3)(x²+4)

Simplify F(x) first.

F(x) = (x²+3 - 2)(x²+4 – 2) / (x²+3)(x²+4)

= ((x² + 3)( x² + 4) -2(x² + 4) -2(x² + 3) + 4) / ((x² + 3) (x² + 4)

= 1 – 2/( x² + 3) -2/( x² + 4) + 4/(( x² +3)( x² + 4))

= 1 – 2/( x² + 3) -2/(x² + 4) + 4 (1/( x² +3) – 1/( x² + 4))

= 1 + 2/( x² + 3) – 6/(x² + 4)

F(x) = 1 + 2/( x² + 3) – 6/(x² + 4)

Integrating F(x), we get

= ∫(1) dx + 2 ∫(1/( x² + (√3)²)) dx - 6∫(1/( x² + 2²)) dx  

= x + 2tan^-1(x/√3)/√3 – 6tan^-1(x/2)/2 + C

Where C is a constant.  

Answer 2

GIVEN:- $\bold{I = \frac{(x^2 + 1)(x^2 + 2)}{(x^2 + 3)(x^2 + 4)}.dx}$

now, $\bold{I = \frac{(x^2 + 3 - 2)(x^2 + 4 - 2)}{(x^2 + 3)(x^2 + 4)}.dx}$

$\bold{I =\frac{(x^2 + 3)(x^2 + 4) - 2(x^2 + 3) - 2(x^2 + 4) + 4}{(x^2 + 3)(x^2 + 4)}.dx}$

$\bold{I = \{1 - 2/(x^2 + 4) - 2/(x^2 + 3) + 4/(x^2 + 3)(x^2 + 4)\}.dx}$

$\bold{I = \{1 - 2/(x^2 + 4) - 2/(x^2 + 3) + 4[1/(x^2 + 3) - 1/(x^2 + 4)]\}.dx}$

[as, we know:- $\bold{dx/(x^2 + a^2) = tan^{-1}x/a}$

$\bold{I = x - 6tan^{-1}x/2 + 2tan^{-1}x/\sqrt{3}}$