[tex]\it a)\ \int^1_0 \dfrac{1}{x+1} dx=ln(x+1) \Big{|}^1_0 =ln2-ln1 =ln2-0=ln2
\\ \\ \\
b)\ \int f^2(x) dx = \int\dfrac{1}{(x+1)^2} dx = \int(x+1)^{-2}dx =-\dfrac{1}{x+1} +\mathcal{C}[/tex]
Relația din enunț devine:
[tex]\it -\dfrac{1}{x+1} \Big{|}^a _0 = \dfrac{1}{2} \Rightarrow -\dfrac{1}{a+1} + 1 =\dfrac{1}{2} \Rightarrow -\dfrac{1}{a+1} = -\dfrac{1}{2} \Rightarrow a=1[/tex]
[tex]\it c)\ \Big{\int} _0 ^1 \dfrac{1}{x^2+1} dx = arctgx \Big{|}_0^1 = arctg1-arctg 0 = \dfrac{\pi}{4} -0=\dfrac{\pi}{4}[/tex]
