How do you integrate #intxsqrt(x^2 + 1) dx#?

1 Answer
Bill K.
Mar 31, 2015

The answer is #\frac{1}{3}(x^{2}+1)^{3/2}+C=\frac{1}{3}(x^{2}+1)\sqrt{x^{2}+1}+C#.

You can either find this answer by doing some educated guessing or by doing a substitution. You can let #u=x^{2}+1# so that #du=2x dx# and your integral becomes #\int\frac{1}{2}u^{1/2}du=\frac{1}{2}\frac{u^{3/2}}{3/2}+C=\frac{1}{2}\cdot\frac{2}{3}u^{3/2}+C=\frac{1}{3}u^{3/2}+C#. Now plug #u=x^{2}+1# to get the same answer as above:

#\int x\sqrt{x^{2}+1}dx=\frac{1}{3}(x^{2}+1)^{3/2}+C=\frac{1}{3}(x^{2}+1)\sqrt{x^{2}+1}+C#

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