0

$$\int \frac{x\ln\left(x+\sqrt{x^{2}+1}\right)}{\sqrt{x^{2}+1}}dx$$

I honestly have been struggling with this integral for 2 hours straight and none of my methods (including substitution and integration by parts) have worked. I desperately need help...

Lorenzo B.
  • 2,252
johny.bravo
  • 87
  • When you used substitution, what function did you try, and how did it turn out? It isn't pretty, but following this up with an appropriate trigonometric substitution and integration by parts yields an integral that can calculated without much difficulty – Clayton Mar 26 '18 at 20:13
  • For instance $x+\sqrt{x^{2}+1}$ which then always left me with x before ln(...) – johny.bravo Mar 26 '18 at 20:17
  • Try $u=\sqrt{x^2+1}$ and note that $\sqrt{u^2-1}=x$. – Clayton Mar 26 '18 at 20:17
  • http://mathworld.wolfram.com/InverseHyperbolicSine.html – Andrew Li Mar 26 '18 at 20:18
  • Hint: $(\sqrt{x^2+1})’=x/\sqrt{x^2+1}$. – Michael Hoppe Mar 26 '18 at 20:18
  • @Clayton this then leaves me with x in the argument of ln – johny.bravo Mar 26 '18 at 20:21
  • What is the source of this integral - what is the motivation to evaluate it? There are an infinite number of integrals that we could randomly look at, and we know most of them won't have a closed form. – Carl Mummert Mar 26 '18 at 20:23
  • @CarlMummert It's from a set my tutor gave me on Mathematical Analysis course at the uni so it sure has a closed form, Wolfram says so too. – johny.bravo Mar 26 '18 at 20:29
  • 2
    Please look at "how to ask a good question" at https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question for advice on posting on this site. You should have indicated the course it was from at the outset, as well as the fact that you knew it had a closed form (and what the closed form was). Posts that merely state a problem with no context are discouraged, as they seem indistinguishable from homework. – Carl Mummert Mar 26 '18 at 20:31
  • Duplicate of this. Also see integration by the hyperbolic substitution. – PinkyWay Jun 17 '21 at 14:56

3 Answers3

1

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

let $u = x+\sqrt{x^{2} + 1}$, then $du = 1+\frac{x}{ \sqrt{x^{2}+1}} dx \implies dx = \frac{\sqrt{x^{2} + 1}}{x + \sqrt{x^{2} + 1}} du$

$$ \int \frac{x \ln(x + \sqrt{x^{2} + 1})}{ \sqrt{x^{2} + 1} } dx = x \int \frac{ \ln(u) }{u} du - \int \left( \int \frac{ \ln(u) }{u} du \right) dx$$

Can you proceed..?

i the end you have to calculate this though ...

$$ \int \ln^{2}(x + \sqrt{x^{2} + 1}) dx $$

Redsbefall
  • 4,845
1

Set $u=x+\sqrt{x^2+1}$ so $$x^2+1=(u-x)^2=u^2-2ux+x^2\implies x=\frac{u^2-1}{2u}$$and $$\frac{du}{dx}=1+\frac{x}{\sqrt{x^2+1}}=\frac{u}{u-x}=\frac{2u^2}{u^2+1}.$$Since $\frac{x}{\sqrt{x^2+1}}=\frac{x}{u-x}=\frac{u^2-1}{u^2+1}$, the integral is $$\int\frac{u^2-1}{2u^2}\ln u du=\frac{1}{2}(\int \ln u du-\int \ln y dy)$$with $y=1/u$, so the result is $$\frac{1}{2}((u+\frac{1}{u})\ln u-u + \frac{1}{u})+C.$$Using $\frac{1}{u}=\sqrt{x^2+1}-x$, we can simplify this to $$\sqrt{x^2+1}\ln (x+\sqrt{x^2+1})-x+C.$$

J.G.
  • 115,835
1

Noting $d\sqrt{x^{2}+1}=\frac{x}{\sqrt{x^2+1}}dx$ and then by integration by parts, one has \begin{eqnarray} &&\int \frac{x\ln\left(x+\sqrt{x^{2}+1}\right)}{\sqrt{x^{2}+1}}dx\\ &=&\int \ln\left(x+\sqrt{x^{2}+1}\right)d\sqrt{x^{2}+1}\\ &=&\sqrt{x^{2}+1}\ln\left(x+\sqrt{x^{2}+1}\right)-\int\sqrt{x^2+1}d\ln\left(x+\sqrt{x^{2}+1}\right)\\ &=&\sqrt{x^{2}+1}\ln\left(x+\sqrt{x^{2}+1}\right)-\int\sqrt{x^2+1}\frac{1}{x+\sqrt{x^{2}+1}}\left(1+\frac{x}{\sqrt{x^2+1}}\right)dx\\ &=&\sqrt{x^{2}+1}\ln\left(x+\sqrt{x^{2}+1}\right)-\int dx\\ &=&\sqrt{x^{2}+1}\ln\left(x+\sqrt{x^{2}+1}\right)-x+C\\ \end{eqnarray}

xpaul
  • 44,000