# What is wrong with this integral reasoning?

$$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx$$ 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.

• Lucas,i doing now a solution manual for a book, and thi kind of integral is hard to me solve
– user245074
Jun 4, 2015 at 16:28
• Also, note that $\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx =\tanh^{-1}\frac{x^2-1}{\sqrt{x^4+1}}$ Apr 23, 2023 at 12:17

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 $\log a = \frac{1}{2}\log a^2.$

• This is now my second-most upvoted answer. I'm... confused. Jun 5, 2015 at 19:29
• Populurity is not in the hardness but cuteness... Apr 23, 2023 at 10:01

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.

• They differ by a constant (0.3465735903...)
– MasB
Jun 4, 2015 at 2:19
• Their derivatives are equal to the integrand, which is what matters. Jun 4, 2015 at 2:21
• The book's answer can be considered the correct one if you just take an appropriate branch cut. In fact, if you put absolute value sign it will work only for real numbers and breaks down completely for contour integrals. Jun 4, 2015 at 5:21
• The whole "absolute value on the log" thing is a bit misleading, so I wish we'd stop teaching that. Yeah, the derivative of $\log |x| + C$ is $x^{-1}$. But the thing is, $x^{-1}$ is undefined at x=0, so really the general antiderivative is $$\int \dfrac{dx}{x} = \begin{cases} \log(x) + C_1 & \text{ if } x > 0, \\ \log(-x) + C_2 & \text{ if } x < 0. \end{cases}$$ at least if you want to work purely in the context of real numbers. Jun 5, 2015 at 4:39

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