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 = I₁ + I₂
$I_{1}=3 \int_{0}^{2} \frac{2 x}{x^{2}+4} d x$
Let x² + 4 = t
2x dx = dt
When x = 0; t = 4
When x = 2; t = 2² + 4 = 8
Substituting t and dt in I₁
$\left.\Rightarrow \mathrm{I}_{1}=3 \int_{4}^{8} \frac{\mathrm{dt}}{\mathrm{t}}=3[\log | \mathrm{t}]\right]_{4}^{8}$ [$\int \frac{1}{x} d x=\log x$]
$\Rightarrow$ I₁ = 3 [log |8| - log |4|] = 3 log $\frac{8}{4}$
$\Rightarrow$ I₁ = 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 = I₁ + I₂
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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