4x-3-x-2-3x-8-dx-

Question Number 31787 by neel1974 last updated on 14/Mar/18

\int \frac{4 x - 3}{x^{2} + 3 x + 8} d x

Answered by sma3l2996 last updated on 14/Mar/18

A = \int \frac{4 x - 3}{x^{2} + 3 x + 8} d x = 2 \int \frac{2 x - 3 / 2}{x^{2} + 3 x + 8} d x

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

= 2 l n (x^{2} + 3 x + 8) - 9 \int \frac{d x}{x^{2} + 3 x + 8} + c

\int \frac{d x}{x^{2} + 3 x + 8} = \int \frac{d x}{x^{2} + 2 \times \frac{3}{2} \times x + {(\frac{3}{2})}^{2} - \frac{9}{4} + 8}

= \int \frac{d x}{{(x + \frac{3}{2})}^{2} + \frac{23}{4}} = \frac{4}{23} \int \frac{d x}{{(\frac{2 x + 3}{\sqrt{23}})}^{2} + 1}

l e t t = \frac{2 x + 3}{\sqrt{23}} \Rightarrow d t = \frac{2}{\sqrt{23}} d x

\int \frac{d x}{x^{2} + 3 x + 8} = \frac{2 \sqrt{23}}{23} \int \frac{d t}{t^{2} + 1} = \frac{2 \sqrt{23}}{23} {t a n}^{- 1} (\frac{2 x + 3}{\sqrt{23}}) + k

s o

A = 2 l n (x^{2} + 3 x + 8) - \frac{18 \sqrt{23}}{23} {t a n}^{- 1} (\frac{2 x + 3}{\sqrt{23}}) + C

Answered by ajfour last updated on 14/Mar/18

I = 2 \int \frac{2 x + 3}{x^{2} + 3 x + 8} d x - 9 \int \frac{d x}{{(x + \frac{3}{2})}^{2} + {(\frac{\sqrt{23}}{2})}^{2}}

I = 2 \ln ∣ x^{2} + 3 x + 8 ∣ - \frac{18}{\sqrt{23}} \tan^{- 1} (\frac{2 x + 3}{\sqrt{23}}) + c .

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