I'm self learning calculus and I stumbled upon the following integral

$\int \frac{\sqrt{x+1} +1}{\sqrt{x+1} -1}dx$

Wolfram couldn't display the result of solving this integral, but I think I solved it. My questions are:

  1. Is long division the best way for solving integrals of this type? Are there any rules of thumb for integrals in general?
  2. Is my solution correct?

My solution:

  1. Long division $\int \frac{\sqrt{x+1} +1}{\sqrt{x+1} -1}dx = \int{(1 + \frac2{\sqrt{x+1}-1})dx} = x+ 2\int{\frac1{\sqrt{x+1}-1}} dx$
  2. Substitute $u=\sqrt{x+1}; du=\frac{1}{2\sqrt{x+1}}dx;\\ x+2\int{\frac1{\sqrt{x+1}-1}dx}=x+4\int{\frac u{u-1}}du$
  3. Long division $x+4\int{\frac u{u-1}}du = x+4\int{(1 + \frac1{u-1})du} = x+4(\sqrt{x+1} +ln(\sqrt{x+1}-1)) + C$
  • $\begingroup$ Multiply numerator and denominator by $\sqrt{x+1}+1$. This holds for all real values of $x$ because $\sqrt{x+1}+1\neq 0,\forall x\in\Bbb R$ $\endgroup$
    – Masacroso
    Feb 5, 2017 at 20:01

1 Answer 1


Yes, your solution is indeed correct.

I get:

$$\int \frac{\sqrt{x+1}+1}{\sqrt{x+1}-1}~dx=x+4\sqrt{x+1}+4\ln(\sqrt{x+1}-1)+C$$

Which is the same solution you've obtained. I did it the same way you have done it, with long division.

Wolfram|Alpha actually does provide the solution. Try these inputs:

Integrate (sqrt(x+1)+1)/(sqrt(x+1)-1) dx

Integrate (sqrt(x+1)+1)/(sqrt(x+1)-1)

Similarly, Wolfram Mathematica works too with this input:

Integrate[(Sqrt[x + 1] + 1)/(Sqrt[x + 1] - 1), x]


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