Question
Evaluate the following integrals:
$\int_{1}^\limits{2}\frac{3\text{x}}{9\text{x}^2-1}\text{ dx}$
$\int_{1}^\limits{2}\frac{3\text{x}}{9\text{x}^2-1}\text{ dx}$
✓
Answer
Let $\text{x}^2=\text{t}$ Then, $2\text{x dx}=\text{dt}$
When $\text{x}=1,\text{t}=1$ and $\text{x}=2,\text{t}=4$
$\therefore\ \text{I}=\int_{1}^\limits{2}\frac{3\text{x}}{9\text{x}^2-1}\text{ dx}$
$\Rightarrow\text{I}=\frac{3}{2}\int_{1}^\limits{4}\frac{\text{dt}}{9\text{t}-1}$
$\Rightarrow\text{I}=\frac{3}{18}\big[\log(9\text{t}-1)\big]^4_1$
$\Rightarrow\text{I}=\frac{3}{18}\big(\log35-\log8\big)$
$\Rightarrow\text{I}=\frac{(\log35-\log8)}{6}$
When $\text{x}=1,\text{t}=1$ and $\text{x}=2,\text{t}=4$
$\therefore\ \text{I}=\int_{1}^\limits{2}\frac{3\text{x}}{9\text{x}^2-1}\text{ dx}$
$\Rightarrow\text{I}=\frac{3}{2}\int_{1}^\limits{4}\frac{\text{dt}}{9\text{t}-1}$
$\Rightarrow\text{I}=\frac{3}{18}\big[\log(9\text{t}-1)\big]^4_1$
$\Rightarrow\text{I}=\frac{3}{18}\big(\log35-\log8\big)$
$\Rightarrow\text{I}=\frac{(\log35-\log8)}{6}$
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