dx-x-2-x-2-4-

Question Number 116815 by bemath last updated on 07/Oct/20

\int \frac{dx}{(x - 2) (x^{2} + 4)} = ?

Answered by john santu last updated on 07/Oct/20

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

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

p u t x = 2 \Rightarrow 1 = 8 A; A = \frac{1}{8}

p u t x = 0 \Rightarrow 1 = 4 . \frac{1}{8} - 2 C; 2 C = - \frac{1}{2}

C = - \frac{1}{4}

p u t x = 1 \Rightarrow 1 = \frac{5}{8} - (B - \frac{1}{2})

B = \frac{5}{8} - \frac{1}{8} = \frac{1}{8}

T h e i n t e g r a l b e c o m e s

I = \int \frac{1}{8 x} d x + \int \frac{\frac{1}{8} x - \frac{1}{4}}{x^{2} + 4} d x

I = \frac{1}{8} \ln ∣ x ∣ + \frac{1}{8} \int \frac{x - 2}{x^{2} + 4} d x

I = \frac{1}{8} \ln ∣ x ∣ + \frac{1}{16} \int \frac{d (x^{2} + 4)}{x^{2} + 4} - \frac{1}{4} \int \frac{d x}{x^{2} + 4}

I = \frac{1}{8} \ln ∣ x ∣ + \frac{1}{16} \ln ∣ x^{2} + 4 ∣ - \frac{1}{8} \tan^{- 1} (\frac{x}{2}) + c

Answered by Dwaipayan Shikari last updated on 07/Oct/20

\int \frac{d x}{(x - 2) (x^{2} + 4)}

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

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

= \frac{1}{8} l o g (x - 2) - \frac{1}{16} l o g (x^{2} + 4) - \frac{1}{8} {t a n}^{- 1} \frac{x}{2} + C