Computing Indefinite Integrals — Question 7

Question 7 - Solution

For $x\ne0$, write $s=\sqrt{1+x^2}$, so $ds=(x/s)\,dx$ and $x^2=s^2-1$.

$$\int\frac{\sqrt{1+x^2}}x\,dx=\int\frac{s^2}{s^2-1}\,ds =s+\frac12\ln\left|\frac{s-1}{s+1}\right|+C.$$

Since $(s-1)(s+1)=x^2$, this becomes

$$\boxed{\sqrt{1+x^2}+\ln\left|\frac{x}{1+\sqrt{1+x^2}}\right|+C.}$$

The formula is valid on either interval $(-\infty,0)$ or $(0,\infty)$; constants may differ between intervals.