1.Let f(x)=∫dx(3+4x2)4−3x2f(x) = \int \frac{dx}{(3+4x^2)\sqrt{4-3x^2}}f(x)=∫(3+4x2)4−3x2dx, ∣x∣<2/3|x|<2/\sqrt3∣x∣<2/3. If f(0)=0f(0)=0f(0)=0 and f(1)=1αβtan−1(α/β)f(1)=\frac{1}{\alpha\beta}\tan^{-1}(\alpha/\beta)f(1)=αβ1tan−1(α/β), α,β>0\alpha,\beta>0α,β>0, then α2+β2=\alpha^2+\beta^2 =α2+β2=Login to continueOnly logged in users canattempt or see the solution.
