Find $\int \frac{\sqrt{x^2+1}(\ln(x^2+1)-2\ln(x)) }{x^4}\,dx$

$$\int \frac{\sqrt{x^2+1}\bigl(\ln(x^2+1)-2\ln(x)\bigr)}{x^4}\,dx$$ This question was asked on a Calculus Exam today, and I didn't manage to solve it. I tried Eliminating sqrt by putting $x^2+1$ as $t$, But it messes everything else. Has anyone got hint how to solve this?

• There was one missing closing parenthesis. I added it after the second logarithm. Please confirm that the expression now in the formula is the correct one. – mickep Oct 21 '15 at 8:04
• Yeah that's right – CuriousSciDude Oct 21 '15 at 9:16

$$\int \frac{\sqrt{x^2+1}(\ln(x^2+1)-2\ln(x)) dx}{x^4}$$ $$=\int \sqrt{(1+\frac{1}{x^2})}\ln(1+\frac{1}{x^2})\cdot \frac{1}{x^3}dx$$ $$=-\int \sqrt{(1+\frac{1}{x^2})}\ln(1+\frac{1}{x^2})d(1+\frac{1}{x^2})$$

Substitute $z^2=1+\frac{1}{x^2}$ and you get $$-\int z\cdot 2\ln z\cdot 2zdz$$ $$=-4I=-4\int z^2 \ln z \,\ dz$$ $$=-4(z^2 \int \ln z-2\int z\cdot (\int \ln z dz)dz)$$ $$=-4(z^2 (z \ln z -z)-2\int z\cdot (z \ln z -z) dz)$$ $$=-4(z^3 \ln z -z^3-2\int (z^2 \ln z -z^2) dz)$$ $$=-4(z^3 \ln z -z^3-2I +\frac{2z^3}{3} )+c$$ $$=-4z^3 \ln z +4z^3+8I -\frac{8z^3}{3}+c$$

Therefore $$-4I=-4z^3 \ln z +4z^3+8I -\frac{8z^3}{3}+c$$ $$-12I=-4z^3 \ln z +4z^3-\frac{8z^3}{3}+c$$ $$-4I=\frac{1}{3}(-4z^3 \ln z +4z^3-\frac{8z^3}{3})+c$$ using $\int \ln z= (z \ln z -z)$

The answer is $$\frac{1}{3}(-4z^3 \ln z +4z^3-\frac{8z^3}{3})+c$$

You need to substitute $z$ back.

• Your integration by parts looks strange to me. – mickep Oct 21 '15 at 8:21
• @mickep Oops sorry I differentiated $\ln x$ instead of integrating !! Correcting it right now. – SchrodingersCat Oct 21 '15 at 8:23
• @mickep Is it okay now? – SchrodingersCat Oct 21 '15 at 8:34
• It would probably have been easier to switched role in your integration by parts: $\int z^2\ln z\,dz=(1/3)z^3\ln z-\int (1/3)z^3\cdot 1/z\,dz$... – mickep Oct 21 '15 at 8:36
• @mickep Yup perhaps. Tried to do it in a different way just. Nothing else. – SchrodingersCat Oct 21 '15 at 8:38

The $\ln x$ term implies that we consider $x>0$ only.

Since you can write $$\log(1+x^2)-2\log x=\log(1+1/x^2)$$ and $$\sqrt{1+x^2}=x\sqrt{1+1/x^2}$$ your integral can be written $$\int\frac{\sqrt{1+1/x^2}\ln(1+1/x^2)}{x^3}\,dx.$$ The substitution $$u=1+1/x^2$$ seems to fit very well. I leave it to you to do the details.

• hmm..you win! +1 – Chinny84 Oct 21 '15 at 8:13
• I hate being the second one posting a very similar answer, so I give you a ↑ for sympathy. – mickep Oct 21 '15 at 8:15
• Ha, I take it however I get it! – Chinny84 Oct 21 '15 at 8:17

$$\frac{\sqrt{x^2+1}\left(\ln\left(x^2+1\right)-2\ln x\right)}{x^4}=\frac{x\sqrt{1+\frac{1}{x^2}}\ln\left(1+\frac{1}{x^2}\right)}{x^4}$$ let set $u=1+\frac{1}{x^2}\implies du = -\frac{2}{x^3}dx$ we can see that your integral becomes $$\int \sqrt{u}\ln u\frac{du}{-2}$$ this can be solved by parts

To depart away from @Mickep answer lets try the sub $x=\tan t$ we find $$\frac{\sqrt{\sec^2 t}\ln \left(\sec^2 t\right)-\ln \left(\tan^2 t\right)}{\tan^4 t}\sec^2 t dt=\frac{\sec^3 t\left(-2\ln \sin t \right)}{\tan^4 t}dt = -2\frac{\ln \sin t}{\sec t \sin^4 t}dt = -\frac{2\cos t \ln \sin t}{\sin^4 t}dt$$ setting $v =\sin t\implies dv = \cos t dt$ we find $$-2\int \frac{\ln v}{v^4} dv$$

Notice, $$\int \frac{\sqrt{x^2+1}\left(\ln(x^2+1)-2\ln x\right)}{x^4}\ dx$$ $$=\int \frac{x\sqrt{1+\frac{1}{x^2}}\left(\ln\left(x^2+1\right)-\ln x^2\right)}{x^4}\ dx$$ $$=\int \frac{\sqrt{1+\frac{1}{x^2}}\ln\left(\frac{x^2+1}{x^2}\right)}{x^3}\ dx$$ $$=\int \sqrt{1+\frac{1}{x^2}}\ln\left(1+\frac{1}{x^2}\right)\frac{dx}{x^3}$$

let $1+\frac{1}{x^2}=t^2\implies \frac{dx}{x^3}=-t\ dt$ $$\int t\ln(t)(-t\ dt)$$$$=-\int t^2\ln(t)\ dt$$ using product rule, $$=\ln(t)\int t^2\ dt-\int \left(\frac{d}{dt}(\ln(t))\int t^2\ dt\right)\ dt$$ $$=\frac{t^3}{3}\ln(t)-\frac{1}{3}\int t^2\ dt$$ $$=\frac{t^3}{3}\ln(t)-\frac{t^3}{9}+C$$ Now, substitute $t=\sqrt{1+\frac{1}{x^2}}$ & simplify