integrate-with-respect-to-x-2x-1-x-2-4x-8-dx-

Question Number 20908 by j.masanja06@gmail.com last updated on 07/Sep/17

i n t e g r a t e w i t h r e s p e c t t o x

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

Answered by Joel577 last updated on 07/Sep/17

I = \int \frac{2 x + 1}{{(x + 2)}^{2} + 4} d x

Let u = x + 2 \to d u = d x

I = \int \frac{2 u - 3}{u^{2} + 4} d u

Let u = 2 \tan θ \to d u = {2 \sec}^{2} θ d θ

\tan θ = \frac{u}{2} \to θ = \tan^{- 1} (\frac{u}{2})

I = \int \frac{4 \tan θ - 3}{4 (\tan^{2} θ + 1)} {2 \sec}^{2} θ d θ

= \int \frac{4 \tan θ - 3}{2} d θ = \int 2 \tan θ - \frac{3}{2} d θ

= - 2 \ln ∣ \cos θ ∣ - \frac{3}{2} θ + C

= - 2 \ln ∣ \frac{2}{\sqrt{u^{2} + 4}} ∣ - \frac{3}{2} \tan^{- 1} (\frac{u}{2}) + C

= - 2 \ln ∣ \frac{2}{\sqrt{{(x + 2)}^{2} + 4}} ∣ - \frac{3}{2} \tan^{- 1} (\frac{x + 2}{2}) + C

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