Question

Ещё 6 интегралов вот так вот

Answer 1

(7)
$\int\limits {\frac{1}{x^2+25}} \, dx = \int\limits {\frac{1}{x^2+5^2}} \, dx =\frac{1}{5}*arctg(\frac{x}{5})+C$

(8)
$\int\limits {(e^x-e^{-2x})} \, dx = \int\limits {e^x} \, dx - \int\limits {e^{-2x}} \, dx =\\\\ = e^x+C - \int\limits {\frac{e^{-2x}}{-2}} \, d(-2x) =\\\\ = e^x+C +\frac{1}{2} \int\limits {e^{-2x}} \, d(-2x) =\\\\ = e^x+C +\frac{1}{2} e^{-2x} =\\\\ = e^x +\frac{1}{2} e^{-2x} +C$

(9)
$\int\limits {\sqrt{1+3x}} \, dx =\frac{1}{3}* \int\limits {\sqrt{1+3x}} \, d(3x) =\frac{1}{3}* \int\limits {\sqrt{1+3x}} \, d(1+3x) =\\\\ =[t=1+3x]=\frac{1}{3}* \int\limits {t^{\frac{1}{2}}} \, dt=\frac{1}{3}*\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+C=\\\\ =\frac{2}{9}*(1+3x)^{\frac{3}{2}}+C$

(10)
$\int\limits {\frac{x}{x^2+1}} \, dx = \int\limits {\frac{1}{x^2+1}} \, d(\frac{x^2}{2}) =\frac{1}{2}*\int\limits {\frac{1}{x^2+1}} \, d(x^2) =\frac{1}{2}*\int\limits {\frac{1}{x^2+1}} \, d(x^2+1) =\\\\ =\frac{1}{2}*ln(x^2+1)+C$

(11)
$\int\limits {sin(x^2)*x} \, dx = \int\limits {sin(x^2)} \, d(\frac{x^2}{2})=\frac{1}{2}*\int\limits {sin(x^2)} \, d(x^2)=\\\\ =\frac{1}{2}*[-cos(x^2)]+C=-\frac{cos(x^2)}{2}+C$

(12)
$\int\limits {\frac{[4-ln(x)]^2}{x}} \, dx = \int\limits {[4-ln(x)]^2*\frac{1}{x}} \, dx =\\\\ = \int\limits {[4-ln(x)]^2} \, d[ln(x)] = -\int\limits {[4-ln(x)]^2} \, d[-ln(x)] =\\\\ = -\int\limits {[4-ln(x)]^2} \, d[4-ln(x)] =-\frac{[4-ln(x)]^3}{3}+C$