What is wrong with this integral reasoning?

Question

$$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx$$ We start by multiplying by $1=\frac{x}{x}$. $$\int\frac{x^{2}+1}{x^{2}\sqrt{x^{4}+1}}xdx$$ Next, we use the substitution $u=x^{2}$;$\frac{du}{2}=xdx$. $$\frac{1}{2}\int\frac{u+1}{u\sqrt{u^{2}+1}}du=\frac{1}{2}\int\frac{1}{\sqrt{u^{2}+1}}du+\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du=$$ $$=\frac{1}{2}\ln\left(u+\sqrt{u^{2}+1}\right)+\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du$$ The problem is now reduced to computing the integral $\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du$. $$\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du=\frac{1}{2}\int\frac{1}{u^2\sqrt{1+\frac{1}{u^{2}}}}du$$ We use substitution again $v=\frac{1}{u}$;$-dv=\frac{1}{u^{2}}du$, then we have the next integral. $$-\frac{1}{2}\int\frac{1}{\sqrt{1+v^{2}}}dv=-\frac{1}{2}\ln\left(v+\sqrt{1+v^{2}}\right)=-\frac{1}{2}\ln\left(\frac{1}{u}+\sqrt{1+\frac{1}{u^{2}}}\right)=$$ $$=-\frac{1}{2}\ln\left(\frac{1+\sqrt{u^{2}+1}}{u}\right)$$ Finally the solution is: $$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx=\frac{1}{2}\ln\left(u+\sqrt{1+u^{2}}\right)-\frac{1}{2}\ln\left(\frac{1+\sqrt{u^{2}+1}}{u}\right)=\frac{1}{2}\ln\left(\frac{x^{4}+x^{2}\sqrt{x^{4}+1}}{1+\sqrt{x^{4}+1}}\right).$$ But the problem is, the book has the following solution: $$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx=\ln\left(\frac{x^{2}-1+\sqrt{x^{4}+1}}{x}\right)$$ which is obviously different.

Answer 1

I have differentiated your answer and the book's answer. They are both equal to the integrand. So your work is indeed correct.

In fact, the book's answer can be considered wrong since it lacks an absolute value around the argument to $ln$, whereas your answer does not need one there since the argument to $ln$ in your expression is strictly positive.

Answer 2

To explicitly check that they differ by a constant, subtract them:

\begin{align*} &\frac{1}{2}\log \frac{x^4+x^2\sqrt{x^4+1}}{1+\sqrt{x^4+1}} - \log \frac{x^2-1+\sqrt{x^4+1}}{x}\\ &= \frac{1}{2} \log \frac{1-x^2+x^4-\sqrt{1+x^4} + x^2\sqrt{1+x^4}}{x^2}\\ &\qquad - \frac{1}{2} \log \frac{2-2x^2+2x^4-2\sqrt{1+x^4}+2x^2\sqrt{1+x^4}}{x^2}\\ &=\frac{1}{2}\log \frac{1}{2}, \end{align*}

where I get the first term by rationalizing the denominator and the second by the identity $\log a = \frac{1}{2}\log a^2.$

Answer 3

The two expressions are equal up to a constant.

Take from your answer $$\frac{x^2(x^2+\sqrt{x^4+1})}{1+\sqrt{x^4+1}}$$ and multiply top and bottom by $\sqrt{x^4+1}-1$ you get

$$\frac{x^4-x^2+1+(x^2-1)\sqrt{x^4+1}}{x^2}=\frac{1}{2}\frac{(x^2-1+\sqrt{x^4-1})^2}{x^2}$$

So $$\frac{1}{2} \ln \frac{x^2(x^2+\sqrt{x^4+1})}{1+\sqrt{x^4+1}}= \frac{1}{2} \ln \frac{1}{2}\frac{(x^2-1+\sqrt{x^4-1})^2}{x^2}=\frac{1}{2}\ln \frac{1}{2}+\frac{1}{2} \ln \frac{(x^2-1+\sqrt{x^4-1})^2}{x^2}= \frac{1}{2}\ln \frac{1}{2}+\ln \frac{x^2-1+\sqrt{x^4-1}}{x}$$