Antiderivitive F(x) of f(x) = $\sqrt{x^2 + 1}$ where F(0) = 5

Question

I did a lot of Googling trying to figure out how to do this, but all the explanations I found had something to do with some kind of ellipse formula that I know nothing about. This is for my Calc II class. Please help, thanks.

Answer 1

Let $$\displaystyle I =\int \sqrt{x^2+1}\cdot 1dx\;,$$ Now using Integration by parts, We get

$$\displaystyle I = \sqrt{x^2+1}\cdot x -\int\frac{x^2}{\sqrt{x^2+1}}dx =\sqrt{x^2+1}\cdot x-\int\frac{\left[(x^2+1)-1\right]}{\sqrt{x^2+1}}dx $$

So we get $$\displaystyle I = \sqrt{x^2+1}\cdot x-\int\sqrt{x^2+1}dx+\int\frac{1}{\sqrt{x^2+1}}dx$$

So we get $$\displaystyle I = \frac{x\cdot\sqrt{x^2+1}}{2}+\frac{1}{2}\cdot \int\frac{1}{\sqrt{x^2+1}}dx$$

Now Put $(x^2+1) = y^2\;,$ Then $$\displaystyle xdx = ydy\Rightarrow \frac{dx}{y} = \frac{dy}{x}$$

so using Ratio and proportion, We get $$\displaystyle \frac{dx}{y} = \frac{dy}{x} = \frac{dx+dy}{y+x}$$

So we get $$\displaystyle I = \frac{x\cdot\sqrt{x^2+1}}{2}+\int\frac{d(x+y)}{(x+y)} = \frac{x\cdot\sqrt{x^2+1}}{2}+\frac{1}{2}\cdot \ln|x+y|+\mathcal{C}$$

So we get $$\displaystyle I = \frac{x\cdot\sqrt{x^2+1}}{2}+\frac{1}{2}\cdot \ln|x+\sqrt{x^2+1}|+\mathcal{C}$$

Now Put $x=0\;,$ and using $f(0) = 5\;,$ We get $\displaystyle 5 = 0+C\Rightarrow \mathcal{C} = 5$

So we get $$\displaystyle F(x) = I = \frac{x\cdot\sqrt{x^2+1}}{2}+\frac{1}{2}\cdot \ln|x+\sqrt{x^2+1}|+5$$

Answer 2

Just make the substitution $x=\tan\theta$ and be prepared to learn how to do $\int \sec^3\theta\,d\theta$.