\[\begin{array}{l}
1)\,\,\int {\frac{{dx}}{{\sqrt {{x^3}} + 2x + \sqrt x }}} = I\,\,\,,x > 0 \\
Let\,\,x = {u^2} \Rightarrow dx = 2u\,du \\
I = \,\int {\frac{{2u}}{{{u^3} + 2{u^2} + u}}} \,du = 2\,\int {\frac{u}{{u{{\left( {u + 1} \right)}^2}}}} \,du = 2\int {{{\left( {u + 1} \right)}^{ - 2}}} du = \frac{{ - 2}}{{u + 1}} + C = \frac{{ - 2}}{{\sqrt x + 1}} + C \\
\\
2)\,\,\int {\frac{{dx}}{{x\ln x + x\sqrt {\ln x} }}} \,\,,x > 1 \\
= 2\int {\frac{1}{{\underbrace {1 + \sqrt {\ln x} }_u}}(\frac{1}{{2x\sqrt {\ln x} }}} \,dx) = \ln \left( {1 + \sqrt {\ln x} } \right) + C \\
3)\,\int {\frac{{dx}}{{{{\cos }^4}x + \frac{1}{2}{{\sin }^2}2x + {{\sin }^4}x}}} \\
\,\, = \,\int {\frac{{dx}}{{{{\cos }^4}x + 2{{\sin }^2}x{{\cos }^2}x + {{\sin }^4}x}}} = \,\int {\frac{{dx}}{{{{\left( {\underbrace {{{\sin }^2}x + {{\cos }^2}x}_{ = 1}} \right)}^2}}}} = \int {dx = x + C} \\
\end{array}\]
; 05-14-2017 08:52 PM
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