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

Question Number 33643 by mondodotto@gmail.com last updated on 21/Apr/18

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

Answered by Joel578 last updated on 21/Apr/18

I = \int \frac{2 x}{x^{2} + 4} d x + 3 \int \frac{1}{x^{2} + 4} d x

u = x^{2} + 4 \to d u = 2 x d x

I = \int \frac{2 x}{u} (\frac{d u}{2 x}) + \frac{3}{2} \tan^{- 1} (\frac{x}{2})

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

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