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Question:

$\int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x$

Solution:

Let $I=\int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x$

$\int \frac{6 x+3}{x^{2}+4} d x=3 \int \frac{2 x+1}{x^{2}+4} d x$

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

$=3 \log \left(x^{2}+4\right)+\frac{3}{2} \tan ^{-1} \frac{x}{2}=\mathrm{F}(x)$

By second fundamental theorem of calculus, we obtain

$I=\mathrm{F}(2)-\mathrm{F}(0)$

$=\left\{3 \log \left(2^{2}+4\right)+\frac{3}{2} \tan ^{-1}\left(\frac{2}{2}\right)\right\}-\left\{3 \log (0+4)+\frac{3}{2} \tan ^{-1}\left(\frac{0}{2}\right)\right\}$

$=3 \log 8+\frac{3}{2} \tan ^{-1} 1-3 \log 4-\frac{3}{2} \tan ^{-1} 0$

$=3 \log 8+\frac{3}{2}\left(\frac{\pi}{4}\right)-3 \log 4-0$

$=3 \log \left(\frac{8}{4}\right)+\frac{3 \pi}{8}$

$=3 \log 2+\frac{3 \pi}{8}$

 

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