Mathematics

Evaluate $$\int _ { 0 } ^ { 2 } \frac { 6 x + 3 } { x ^ { 2 } + 4 } d x$$


SOLUTION

$$\\\int_{0}^{2}(\frac{6x+3}{x^2+4})dx\\=3\int_{0}^{2}(\frac{2x}{x^2+4})dx+3\int_{0}^{2}(\frac{1}{x^2+2^2})dx\\=3[log(x^2+4)]_{0}^{2}+3\times(\frac{1}{2})\left[tan^{-1}(\frac{x}{2})\right]_{0}^{2}\\=3(log8-log4)+(\frac{3}{2})(tan^{-1}1-tan^{-1}0)\\=3log2+(\frac{3}{2})\times(\frac{\pi}{4})\\=3log2+(\frac{3\pi}{8})$$

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Subjective Medium Published on 17th 09, 2020
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