Question

Evaluate the definite integrals: $\int^3_2 {\frac{x\ dx}{x^2 +1}$

Answer 1

Answer:

$log\sqrt{2}$

Step-by-step explanation:

$Formula\: used:\\\\\int{\frac{f'(x)}{f(x)}}\:dx=log[f(x)]+c\\\\Now,\\\\\int\limits^3_2{\frac{x}{x^2+1}}\:dx\\\\=\frac{1}{2}\int\limits^3_2{\frac{2x}{x^2+1}}\:dx\\\\=\frac{1}{2}[log{(x^2+1)}]^3_2\\\\=\frac{1}{2}[log{(3^2+1)}-log{(2^2+1)} ]\\\\=\frac{1}{2}[log10-log5]\\\\=\frac{1}{2}[log\frac{10}{5}]\\\\=\frac{1}{2}[log2]\\\\=log2^\frac{1}{2}\\\\=log\sqrt{2}$