Question 512 Marks
Evaluate the following integrals:
$\int\limits^1_0\frac{2\text{x}}{1+\text{x}^2}\text{ dx}$
$\int\limits^1_0\frac{2\text{x}}{1+\text{x}^2}\text{ dx}$
Answer
View full question & answer→We have,
$\text{I}=\int\limits^1_0\frac{2\text{x}}{1+\text{x}^2}\text{ dx}$
Putting $1+\text{x}^2=\text{t}$
$2\text{x dx}=\text{dt}$
When $\text{x}\rightarrow0;\text{ t}\rightarrow1$
And $\text{x}\rightarrow1;\text{ t}\rightarrow2$
$\therefore\ \text{I}=\int\limits^1_0\frac{\text{dt}}{\text{t}^2}$
$=\Big[\log_{\text{e}}|\text{t}|\Big]^2_1$
$=\log_{\text{e}}2-\log_{\text{e}}1$
$=\log_{\text{e}}2-0$
$=\log_{\text{e}}2$
$\text{I}=\int\limits^1_0\frac{2\text{x}}{1+\text{x}^2}\text{ dx}$
Putting $1+\text{x}^2=\text{t}$
$2\text{x dx}=\text{dt}$
When $\text{x}\rightarrow0;\text{ t}\rightarrow1$
And $\text{x}\rightarrow1;\text{ t}\rightarrow2$
$\therefore\ \text{I}=\int\limits^1_0\frac{\text{dt}}{\text{t}^2}$
$=\Big[\log_{\text{e}}|\text{t}|\Big]^2_1$
$=\log_{\text{e}}2-\log_{\text{e}}1$
$=\log_{\text{e}}2-0$
$=\log_{\text{e}}2$