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)) ∴I=x2(cos(lnx)+sin(lnx))+c
None of the above
∫dxxn(√ax+b)=−√ax+b(n−1)bxn−1−A∫dxxn−1(√ax+b)+C
⇒∫1+Axxn(√ax+b)dx=−√ax+b(n−1)bxn−1+C
Differentiating on both sides,
1+Axxn(√ax+b)=−xn−1a2(ax+b)−12−(ax+b)12(n−1)xn−2b(n−1)x2n−2
⇒1+Axxn√ax+b=−ax−2(n−1)(ax+b)b(n−1)2xn√ax+b
Simplifying further, it gives
A=2n−32n−2ab
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