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

1 Answer
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