find-ln-x-x-2-x-2-dx-

Question Number 36430 by prof Abdo imad last updated on 02/Jun/18

f i n d \int \frac{l n (x + x^{2})}{x^{2}} d x

Commented by abdo mathsup last updated on 03/Jun/18

l e t i n t e g r a t e b y p a r t s u^{^{'}} = \frac{1}{x^{2}} a n d v = l n (x + x^{2})

I = - \frac{1}{x} l n (x + x^{2}) + \int \frac{1}{x} \frac{2 x + 1}{x + x^{2}} d x

= - \frac{l n (x + x^{2})}{x} + \int \frac{2 x + 1}{x^{2} (x + 1)} d x b u t

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

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

F (x) = \frac{1}{x^{2} (x + 1)} = \frac{a}{x} + \frac{b}{x^{2}} + \frac{c}{x + 1}

b = {l i m}_{x \to 0} x^{2} F (x) = 1

c = {l i m}_{x \to - 1} (x + 1) F (x) = 1 \Rightarrow

F (x) = \frac{1}{x} + \frac{b}{x^{2}} + \frac{1}{x + 1}

F (1) = \frac{1}{2} = 1 + b + \frac{1}{2} \Rightarrow b = - 1 \Rightarrow

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

\int F (x) d x = l n ∣ x ∣ + \frac{1}{x} - l n ∣ x + 1 ∣ + c \Rightarrow

I = - \frac{l n (x^{2} + x)}{x} - \frac{2}{x} - l n ∣ x ∣ - \frac{1}{x} + l n ∣ x + 1 ∣ + c \Rightarrow

I = - \frac{l n (x^{2} + x)}{x} - \frac{3}{x} - l n ∣ x ∣ + l n ∣ x + 1 ∣ + c .

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18

l n (x + x^{2}) \times \frac{- 1}{x} - \int \frac{1}{x + x^{2}} \times (1 + 2 x) \times \frac{- 1}{x} d x

\frac{- l n (x + x^{2})}{x} + \int \frac{1 + 2 x}{x^{2} (1 + x)} d x

I_{2} = \int \frac{1 + 2 x}{x^{2} (1 + x)} d x

= \int \frac{1 + x + x}{x^{2} (1 + x)} d x

= \int \frac{d x}{x^{2}} + \int \frac{d x}{x (1 + x)}

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

= \frac{- 1}{x} + \int \frac{d x}{x} - \int \frac{d x}{1 + x}

= \frac{- 1}{x} + l n (\frac{x}{1 + x})

= - \frac{l n (x + x^{2})}{x} - \frac{1}{x} + l n \frac{x}{1 + x}