Explanation
Let I=∫cos(lnx)dx I=xcos(lnx)+∫xsin(lnx)1xdx ⇒I=xcos(lnx)+∫sin(lnx)dx =xcos(lnx)+[xsin(lnx)−∫xcos(lnx).1xdx] ⇒2I=x(cos(lnx)+sin(lnx)) ∴
None of the above
\displaystyle\int {\frac{dx}{ { x }^{ n }\left( \sqrt { ax+b } \right) } } = -\frac { \sqrt { ax+b } }{ \left( n-1 \right) b{ x }^{ n-1 } } -A\int { \frac { dx }{ { x }^{ n-1 }\left( \sqrt { ax+b } \right) } } + C
\Rightarrow\displaystyle \int { \frac { 1+Ax }{ { x }^{ n }\left( \sqrt { ax+b } \right) } } dx = -\frac { \sqrt { ax+b } }{ \left( n-1 \right) b{ x }^{ n-1 } } + C
Differentiating on both sides,
\dfrac{1+Ax}{x^n(\sqrt{ax+b})} =- \dfrac{x^{n-1}\dfrac{a}{2}(ax+b)^{-\frac12}-(ax+b)^{\frac12}(n-1)x^{n-2}}{b(n-1)x^{2n-2}}
\Rightarrow\dfrac{1+Ax}{x^n\sqrt{ax+b}}=-\dfrac{ax-2(n-1)(ax+b)}{b(n-1)2x^n\sqrt{ax+b}}
Simplifying further, it gives
A = \dfrac { 2n-3 }{ 2n-2 } \dfrac { a }{ b }
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