Question

ПОМОГИТЕ ПОЖАЛУЙСТА 2 ИНТЕГРАЛА

Answer 1

(9)
$\int\limits^1_0 {\frac{x}{(x^2+1)^2}} \, dx = \frac{1}{2}*\int\limits^1_0 {\frac{1}{(x^2+1)^2}} \, d(x^2) =\frac{1}{2}*\int\limits^1_0 {\frac{1}{(x^2+1)^2}} \, d(x^2+1) =\\\\ =\frac{1}{2}*\int\limits^{1^2+1}_{0^2+1} {t^{-2}}} \, dt =\frac{1}{2}*\int\limits^{2}_{1} {t^{-2}}} \, dt =\frac{1}{2}*\frac{t^{-1}}{-1}|^{2}_{1}=-\frac{1}{2}*\frac{1}{t}|^2_1=\\\\= -\frac{1}{2}*[\frac{1}{2}-\frac{1}{1}]=\frac{1}{4}$

(10)
$\int\limits^8_0 {\frac{1}{\sqrt{3x+1}}} \, dx = \frac{1}{3}*\int\limits^8_0 {\frac{1}{\sqrt{3x+1}}} \, d(3x+1) =\frac{1}{3}*\int\limits^{3*8+1}_{3*0+1} {\frac{1}{\sqrt{t}}} \, dt =\\\\ =\frac{1}{3}* \int\limits^{25}_{1} {t^{-\frac{1}{2}}} \, dt=\frac{1}{3}*\frac{t^{\frac{1}{2}}}{\frac{1}{2}}|^{25}_1=\frac{2}{3}*\sqrt{t}|^{25}_3=\\\\ =\frac{2}{3}*[\sqrt{25}-\sqrt{1}]=\frac{2}{3}*[5-1]=\frac{8}{3}$