Explanation
{y^2}dx + \left( {{x^2} - xy + 42} \right)dy = 0
\displaystyle {{dx} \over {dy}} = {{ - \left( {{x^2} - xy + {y^2}} \right)} \over {{y^2}}}
\displaystyle {{dx} \over {dy}} = - {{{x^2}} \over {{y^2}}} + {x \over y} - 1
Put v = x/y\implies x=vy
dx/dy = v + ydv/dy
\Rightarrow v + ydv/dy = - {v^2} + v - 1
\Rightarrow ydv/dy = - \left( {{v^2} + 1} \right)
\displaystyle {{dv} \over {{v^2} + 1}} = - {1 \over y}dy
Integrating both side
\Rightarrow {\tan ^{ - 1}}v = - \log y + c
\Rightarrow {\tan ^{ - 1}}x/y = - \log y + c\,\,\,\,c = - c
\Rightarrow {\tan ^{ - 1}}\left( {x/y} \right) + \log y + c = 0
\begin{matrix} \dfrac { { dy } }{ { dx } } =\dfrac { g }{ n } +\tan \left( { \dfrac { y }{ x } } \right) \\ It\, \, is\, \, { { homogenouous\ diffenrential\ equation } } \\ { { Lety= } }\sqrt { x } \\ \Rightarrow \dfrac { { dy } }{ { dx } } =\sqrt { } +\dfrac { { xdv } }{ { dx } } \\ and\, \, put\, \, in\, \, differential\, \, equation\, \, \\ \Rightarrow \sqrt { } +x\dfrac { { dv } }{ { dx } } =\sqrt { } +\tan v \\ \Rightarrow \dfrac { { xdv } }{ { dx } } =\tan v \\ \Rightarrow \int { \dfrac { { dv } }{ { \tan v } } =\int { \dfrac { { dx } }{ x } } } \\ \Rightarrow \int { \sec v=\log x+c } \\ \Rightarrow \log \left| { \sec v=\tan v } \right| \log xc \\ \Rightarrow \log \left| { \sec \dfrac { y }{ x } +\tan \dfrac { y }{ x } } \right| =\log xc \\ \, \, \, \left| { \sec \dfrac { y }{ x } +\tan \dfrac { y }{ x } } \right| =xc\, \, \, Ans. \\ \end{matrix}
Consider the given integral.
I=\int{{{\left( {{x}^{x}} \right)}^{x}}\left( 2x\log x+x \right)}dx
I=\int{\left( {{x}^{{{x}^{2}}}} \right)\left( 2x\log x+x \right)}dx
Let t={{x}^{{{x}^{2}}}}
\log t={{x}^{2}}\log x
\dfrac{1}{t}\dfrac{dt}{dx}=\left( 2x\log x+x \right)
dt={{x}^{{{x}^{2}}}}\left( 2x\log x+x \right)dx
Therefore,
I=\int{1dt}
I=t+C
On putting the value of t, we get
I={{x}^{{{x}^{2}}}}+C
I={{x}^{{{x}^{x}}}}+C
Hence, this is the answer.
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