find-1-1-x-2-arctan-2x-dx-

Question Number 86956 by abdomathmax last updated on 01/Apr/20

f i n d \int (1 - \frac{1}{x^{2}}) a r c t a n (2 x) d x

Commented by Ar Brandon last updated on 01/Apr/20

L e t u = a r c t a n (2 x) \Rightarrow d u = 2 \cdot \frac{1}{1 + {(2 x)}^{2}} d x

d v = (1 - \frac{1}{x^{2}}) d x \Rightarrow v = (x + \frac{1}{x})

I = (x + \frac{1}{x^{2}}) a r c t a n (2 x) - 2 \int (x + \frac{1}{x}) \cdot (\frac{1}{1 + 4 x^{2}}) d x

= (x + \frac{1}{x^{2}}) a r c t a n (2 x) - 2 \int \frac{x^{2} + 1}{x} \cdot \frac{1}{1 + 4 x^{2}} d x

= (x + \frac{1}{x^{2}}) a r c t a n (2 x) - 2 \int \frac{x^{2} + 1}{x (1 + 4 x^{2})} d x

\frac{x^{2} + 1}{x (1 + 4 x^{2})} = \frac{A}{x} + \frac{B x + C}{4 x^{2} + 1}

= \frac{A (4 x^{2} + 1) + (B x + C) x}{x (4 x^{2} + 1)} = \frac{(4 A + B) x^{2} + C x + A}{x (4 x^{2} + 1)}

4 A + B = 1

C = 0

A = 1 \Rightarrow B = - 3

\int \frac{x^{2}}{x (1 + 4 x^{2})} d x = \int (\frac{1}{x} - \frac{3 x}{1 + 4 x^{2}}) = \int \frac{1}{x} d x - \frac{3}{8} \int \frac{8 x}{1 + 4 x^{2}} d x

= l n x - \frac{3}{8} l n (1 + 4 x^{2}) + C o n s t a n t

\int (1 - \frac{1}{x^{2}}) a r c t a n (2 x) d x = (x + \frac{1}{x}) a r c t a n (2 x) - 2 l n x + \frac{3}{4} l n (1 + 4 x^{2}) + c o n s t a n t

Commented by mathmax by abdo last updated on 01/Apr/20

t h a n k y o u s i r

Commented by Ar Brandon last updated on 01/Apr/20

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