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Question Number 68767 by aliesam last updated on 15/Sep/19
Commented by ~ À ® @ 237 ~ last updated on 15/Sep/19
Letnameditfn(x)Letnamed∀a>0g(a,x)=∫dxa+x2Sog(a,x)=1a∫dx1+x2a=1a∫d(xa)1+(xa)2g(a,x)=arctan(xa)a+cNowjustascertainthat∂g(a,x)∂a=−∫dx(a+x2)2∂2g(a,x)∂a2=∫2dx(a+x2)3......∂ng(a,x)∂an=∫(−1)n(n−1)(a+x2)ndxFinally∫dx(a+x2)n=(−1)nn−1∂n−1g(a,x)∂an−1fn(x)=(−1)nn−1∂n−1g∂an−1(1,x)
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