How can we solve this integration? $$\int \frac{1+\sqrt{1+x^2}}{x^2+\sqrt{1+x^2}}dx$$ I tried to make the following substitution $$1+x^2=w^2$$ but this substitution complicated the integral.
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1$\begingroup$ Wolfram gives this monster of an answer. $\endgroup$ – Peter Foreman Oct 24 '19 at 0:17
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Let $x=\tan u$
$$\int \frac{\sec^2u+\sec^3u}{\tan^2u+\sec u}$$
$$\int \frac{\sec^2u+\sec u \tan^2 u +\sec u}{\tan^2u+\sec u}$$
First two terms cancel with the denominator the third term and the denominator think about it
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Like the answer of @AmerYR, taking $1+x^2=u^2$ $$I=\int \frac{1+\sqrt{1+x^2}}{x^2+\sqrt{1+x^2}} dx= \int \sec u ~du + \int \frac{sec u}{\tan^2 u+ sec u} du$$ $$ \implies I =\log (\tan u+ sec~ u)+J(u)$$ Next, using $\sin u=\frac{2 \tan(u/2)}{1+\tan^(u/2)},~~ \cos u= \frac{1-\tan^(u/2)}{1+\tan^2(u/2)}$ and $t=\tan(u/2), dt= sec^2 (u/2) du/2$, we get
$$J=\int \frac{\cos u ~du} {\sin^2 u + \cos u}= \frac{(1-t^2)/(1+t^2)}{4t^2/(1+t^2)^2+(1-t^2)/(1+t^2)} \frac{2dt}{(1+t^2)}$$ $$\implies J= 2 \int \frac{t^2-1}{t^4-4t^2-1} dt= 2 \int \frac{t^2-1}{(t^2+a^2)(t^2-b^2)} dt = 2 \int \left(\frac{A}{t^2-a^2} + \frac{B}{t^2+b^2} \right) dt$$ $$\implies 2 \left(\frac{A}{2a} \log \frac{t-a}{t+a} + \frac{B}{b} \tan^{-1} \frac{t}{b} \right) +C$$ Here $a=\sqrt{2+\sqrt{5}}, b=\sqrt{\sqrt{5}-2}$, $A=\frac{\sqrt{5}+1}{2\sqrt{5}}$, $B=\frac{\sqrt{5}-1}{2\sqrt{5}}.$
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Substitute $x=\sinh u$,
$$I=\int \frac{1+\sqrt{1+x^2}}{x^2+\sqrt{1+x^2}}dx= \int \frac{\cosh u+\cosh^2 u}{\cosh^2 u+\cosh u - 1}du = u+I_1$$
where,
$$I_1=\int \frac{du}{\cosh^2 u+\cosh u-1}$$
Next, use the substitution $\cosh u = \frac{1+t^2}{1-t^2}$, along with $d u = \frac{2dt}{1-t^2}$,
$$I_1=2\int \frac{t^2-1}{t^4-4t^2-1}dt$$ $$=\frac{1}{\sqrt5}\int\left( \frac{\sqrt5+1}{t^2+\sqrt5-2}+\frac{\sqrt5-1}{t^2-\sqrt5-2}\right)dt$$ $$=\sqrt{\frac25}\sqrt{\sqrt5+1}\tan^{-1}\left(\sqrt{\sqrt5+2}\>t\right) - \sqrt{\frac25}\sqrt{\sqrt5-1}\tanh^{-1}\left(\sqrt{\sqrt5-2}\>t\right)$$
Thus, the solution is,
$$I= \sinh^{-1}x + \sqrt{\frac25(\sqrt5+1)}\tan^{-1}\left(\sqrt{\sqrt5+2}\>t\right) -\sqrt{\frac25(\sqrt5-1)}\tanh^{-1}\left(\sqrt{\sqrt5-2}\>t\right)$$
where,
$$t= \left(\frac{\cosh u-1}{\cosh u+1}\right)^{\frac12} = \left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right)^{\frac12}$$
