Let I = ∫x^2/(x^2+a^2)(x^2+b^2)dx Now,x^2/(x^2+a^2)(x^2+b^2) [let x^2 = t] = t/(t+a^2)(t+b^2)=A/(t+a^2)+B/(t+b^2) t=A(t+b^2)+B(t+a^2) On comparing the c ...

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Integration Using Substitution

∫x2/x2+a2x2+b...

Question

$\int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x =$

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Solution

Let I = $\int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x N o w, \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} [l e t x^{2} = t] = \frac{t}{(t + a^{2}) (t + b^{2})} = \frac{A}{(t + a^{2})} + \frac{B}{(t + b^{2})} t = A (t + b^{2}) + B (t + a^{2}) O n c o m p a r i n g t h e c o e f f i c i e n t s o f t, w e g e t . A + B = 1 b^{2} A + a^{2} B = 0 A + B = 1 b^{2} A + a^{2} B = 0 \Rightarrow b^{(} 1 - B) + a^{2} B = 0 \Rightarrow b^{2} - b^{2} B + a^{2} B = 0 \Rightarrow b^{2} + (a^{2} - b^{2}) B = 0 \Rightarrow B = \frac{- b^{2}}{a^{2} - b^{2}} = \frac{b^{2}}{b^{2} - a^{2}} F r o m E q . (i) . A + \frac{b^{2}}{b^{2} - a^{2}} = 1 \Rightarrow A = \frac{b^{2} - a^{2} - b^{2}}{b^{2} - a^{2}} = \frac{- a^{2}}{b^{2} - a^{2}} = \frac{- a^{2}}{(b^{2} - a^{2})} \int \frac{1}{x^{2} + a^{2}} d x + \frac{b^{2}}{b^{2} - a^{2}} \int \frac{1}{x^{2} + b^{2}} d x = \frac{- a^{2}}{b^{2} - a^{2}} . \frac{1}{a} t a n^{- 1} \frac{x}{a} + \frac{b^{2}}{b^{2} - a^{2}} . \frac{1}{b} t a n^{- 1} \frac{x}{b} = \frac{1}{b^{2} - a^{2}} [- a t a n^{- 1} \frac{x}{a} + b t a n^{- 1} \frac{x}{b}] = \frac{1}{b^{2} - a^{2}} [a t a n^{- 1} \frac{x}{a} - b t a n^{- 1} \frac{x}{b}]$

Suggest Corrections

∫x2(x2+a2)(x2+b2)dx=

$\int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x =$