Question

va rog sa ma ajutati

Answer 1

...........................................................

Answer 2

$\displaystyle \mathtt{\int\limits _2^4 \frac{x^2-1}{x+1}dx= \int\limits _2^4\frac{(x-1)(x+1)}{x+1}dx=\int\limits_2^4(x-1)dx=\int\limits_2^4xdx-\int\limits_2^41dx=}\\ \\ \mathtt{= \frac{x^2}{2}\Bigg|_2^4-x\Bigg|_2^4=\left( \frac{4^2}{2}- \frac{2^2}{2} \right)-(4-2)= \frac{16-4}{2} -2= \frac{12}{2}-2=}\\ \\ \mathtt{=6-2=4}\\ \\ \mathtt{\int\limits _2^4 \frac{x^2-1}{x+1}dx=4}$