Question

Evaluate the definite integral: $∫^3_2\frac{xdx}{x^2+1}$

Answer 1

$Let I=∫^3_2\frac{x}{x^2+1}dx$

$6∫\frac{x}{x^2+1}dx=\frac{1}{2}∫\frac{2x}{x^2+1}dx=\frac{1}{2}log(1+x^2)=F(x)$

By second fundamental theorem of calculus,we obtain

$I=F(3)-F(2)$

$=\frac{1}{2}[log(1+(3)^2)-log(1+(2)^2)]$

$=\frac{1}{2}[log(10)-log(5)]$

$=\frac{1}{2}log(\frac{10}{5})=\frac{1}{2}log2$