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1.17 Integrating term (b)

b) For the second term we make the substitution $x=\alpha\tan t$ . Then $\frac{dx}{dt}={\alpha\sec^{2}t}$ and

x^{2}+\alpha^{2}={\alpha^{2}\tan^{2}t+\alpha^{2}=}\,{\alpha^{2}(1+\tan^{2}t)=% \alpha^{2}\sec^{2}t}

so the integral of the second term is:

\int\frac{K\,dx}{x^{2}+\alpha^{2}}=\int\frac{K\frac{dx}{dt}\,dt}{x^{2}+\alpha^% {2}}={\int\frac{K\alpha\sec^{2}t\,dt}{\alpha^{2}\sec^{2}t}=}{\frac{Kt}{\alpha}% +C}

which, in terms of $x$ , equals $\frac{K}{\alpha}\tan^{-1}\left(\frac{x}{\alpha}\right)$ .

Putting (a) and (b) together, we obtain:

\int\frac{Hx+K}{x^{2}+\alpha^{2}}dx={\frac{H}{2}\log(x^{2}+\alpha^{2})+\frac{K% }{\alpha}\tan^{-1}\left(\frac{x}{\alpha}\right)+C.}

For details of how to integrate (ii) in general, see the workshop exercises.