Integration and derivative together

Question

I have the function $f(x)=\ln(x+\sqrt{x^2+1})$ and I have already find that $$f'(x)=\frac{1}{\sqrt{x^2+1}}$$ and I want to find the $$\int_0^1{\sqrt{x^2+1}}dx$$ so: $$\int_0^1{\sqrt{x^2+1}}dx=\int_0^1{\frac{1}{f'(x)}}dx$$ but I am stack and I don't know how to continue. Any ideas?

Answer 1

You need to use a trick:

$$\int\sqrt{x^2+1}\,dx=\int\frac{x^2+1}{\sqrt{x^2+1}}dx=\int\frac{x^2}{\sqrt{x^2+1}}dx+\int\frac1{\sqrt{x^2+1}}dx$$ and your $f'(x)$ appears.

But the other term is still a pain in the neck. You can address it by parts,

$$\int x\frac x{\sqrt{x^2+1}}dx=x\sqrt{x^2+1}-\int\sqrt{x^2+1}\,dx$$ and... you get back to where you started from. But before falling into despair, if you put it all together, you get

$$I=x\sqrt{x^2+1}-I+f(x)+C,$$ from which you draw $I$.

Answer 2

Write $$\int_0^1{\sqrt{x^2+1}}\,dx=\int_0^1\frac{x^2+1}{\sqrt{x^2+1}}\,dx$$ and use integration by parts with $$f(x)=x^2+1\implies f'(x)=2x$$ and $$g'(x)=\dfrac1{\sqrt{x^2+1}}\implies g(x)=\ln(x+\sqrt{x^2+1})$$ so $$\begin{align}\int_0^1{\sqrt{x^2+1}}\,dx&=\left[(x^2+1)\ln(x+\sqrt{x^2+1})\right]_0^1-\int_0^1 2x\ln(x+\sqrt{x^2+1})\,dx\\&=2\ln(1+\sqrt2)-2\int_0^1x\sinh^{-1}x\,dx\end{align}$$

Answer 3

Do an integration by parts. Let $I = \int \sqrt{x^2 + 1}\;dx$. Then \begin{align} I &= x\sqrt{x^2+1} - \int\frac{x^2}{\sqrt{x^2+1}}\;dx \\ &= x\sqrt{x^2+1} - \int\frac{x^2+1-1}{\sqrt{x^2+1}}\;dx \\ &= x\sqrt{x^2+1} - I + \int\frac{1}{\sqrt{x^2+1}}\;dx \end{align} So the integral $I$ is: $$I = \frac{1}{2}\left(x\sqrt{x^2+1} + \log(x + \sqrt{x^2+1})\right)$$ Now, just plug in your integration limits!

EDIT: If you plug in your limits, you should get: $$\int_0^1\sqrt{x^2+1}\;dx = \frac{1}{2}\left(\sqrt{2} + \log(1+\sqrt{2})\right)$$

Answer 4

Put $$x=\tan\theta$$$$dx={d\theta}\,{\sec^2\theta}$$