# Evaluate $\int\left(\sqrt{4-x^2}+x\right)dx$

I'm looking for simple way to solve

$$\int\left(\sqrt{4-x^2}+x\right) \, dx$$

I tried substitute $x=2\sin u$ and then

$$\cdots =\frac{x^2}{2}+\frac 1 2x\sqrt{4-x^2}+2\arcsin (x/2)+c$$

I'm looking for other solution please

• I think what you tried is the most elementary! Commented Aug 1, 2016 at 20:08
• "Solve" is not the right word here. "Evaluate" is. ("Solve" is one of those words that people not very familiar with mathematical terminology use as a catch-all when they don't know what word to use. It has legitimate uses as well.) $\qquad$ Commented Aug 1, 2016 at 20:09
• Partial integration, you get the original integral back plus a term proportional to the derivative of an arcsin. You can then bring the square root term to the other side and solve for the integral of it. Commented Aug 1, 2016 at 20:10
• Do you know that $\int \frac{dt}{\sqrt{1-t^2}}=\arcsin t+C$? If so one can use an integration by parts argument. There is also a purely geometric argument, involving area of part of a circle. Commented Aug 1, 2016 at 20:11

Another way forward is to integrate by parts with $u=\sqrt{4-x^2}$ and $v=x$. Then, we have

\begin{align} \int \sqrt{4-x^2}\,dx&=x\sqrt{4-x^2}+\int \frac{x^2}{\sqrt{4-x^2}}\,dx\\\\ &=x\sqrt{4-x^2}+\int \frac{x^2-4+4}{\sqrt{4-x^2}}\,dx\\\\ &=x\sqrt{4-x^2}-\int \sqrt{4-x^2}\,dx+4\int \frac{1}{\sqrt{4-x^2}}\,dx\\\\ 2 \int \sqrt{4-x^2}\,dx&=x\sqrt{4-x^2}+4\int \frac{1}{\sqrt{4-x^2}}\,dx\\\\ \int \sqrt{4-x^2}\,dx&=\frac12 x\sqrt{4-x^2}+2\int \frac{1}{\sqrt{4-x^2}}\,dx\\\\ &=\frac12 x\sqrt{4-x^2}+2\int \frac{1}{\sqrt{1-(x/2)^2}}\,d(x/2)\\\\ &=\frac12 x\sqrt{4-x^2}+2\arcsin(x/2)+C \end{align}