Evaluate the definite integral _ 0 ^ 2 6 x+3 x^ 2 +4 d x

Question

Evaluate the definite integral $\int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x$

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Answer

Let $I=\int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x$
$I=3 \int_{0}^{2} \frac{2 x+1}{x^{2}+4}=3 \int_{0}^{2} \frac{2 x}{x^{2}+4} d x+3 \int_{0}^{2} \frac{1}{x^{2}+4} d x$
$\therefore I = I_1 + I_{2 }$
$I_{1}=3 \int_{0}^{2} \frac{2 x}{x^{2}+4} d x$
Let $x^2 + 4 = t$
$2x dx = dt$
When $x = 0; t = 4$
When $x = 2; t = 2^2 + 4 = 8$
Substituting t and dt in $I_{1 }$
$\left.\Rightarrow \mathrm{I}_{1}=3 \int_{4}^{8} \frac{\mathrm{dt}}{\mathrm{t}}=3[\log | \mathrm{t}]\right]_{4}^{8} [\int \frac{1}{x} d x=\log x]$
$\Rightarrow I_1 = 3 [\log |8| - \log |4|] = 3 \log \frac{8}{4}$
$\Rightarrow I_1 = 3 \log |2|$
$\mathrm{I}_{2}=3 \int_{0}^{2} \frac{1}{\mathrm{x}^{2}+4} \mathrm{dx}=3 \int_{0}^{2} \frac{1}{\mathrm{x}^{2}+2^{2}} \mathrm{d} \mathrm{x} [\frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c]$
$\Rightarrow I_{2}=3 \times \frac{1}{2}\left[\tan ^{-1} \frac{x}{2}\right]_{0}^{2} = \frac{3}{2}\left[\tan ^{-1} \frac{2}{2}-\tan ^{-1} \frac{0}{2}\right] = \frac{3}{2}\left[\tan ^{-1} 1-\tan ^{-1} 0\right]$
$\Rightarrow I_{2}=\frac{3}{2} \times \frac{\pi}{4}=\frac{3 \pi}{ 8}$
Now $I = I_1 + I_{2 }$
$I = 3 \log 2 + \frac{3\pi}{8}$
$\int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x=3 \log 2+\frac{3 \pi}{8}$

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