# Defined integral of a function

I am a student, and I was assigned this integral as homework:

$$\int_0^{2\sqrt{2}} \frac{x}{\sqrt{1+x^2}} \, dx$$

• Hint: compute the derivative of the denominator. – Raymond Manzoni Dec 27 '12 at 17:48
• (+1) for your question. Should we downvote the easy questions? Each questions has its value. – user 1357113 Dec 27 '12 at 18:30

This is an easy and elementary definite integral. First of all consider $\int\frac{x}{\sqrt{1+x^2}} \, dx$. You can solve it by taking $u=1+x^2$ inside the radical and use it as a good substitution. Then $$u=1+x^2\longrightarrow du=2x \, dx\longrightarrow x \, dx=\frac{du}{2}$$ so the indifinte integral would be $$\int\frac{du/2}{\sqrt{u}}.$$ I think you can handle it.

• Yes I can, thank you very much for your hint; my problem was that I tried to take the square as "u". – Flora Dec 27 '12 at 18:04
• @Snowzurfer: Note that after doing the indefinite integral, you should do what my friend amWhy did for you. +1 for his completing answer. – mrs Dec 27 '12 at 18:07
• @@Babak Sorouh: for the well explained way. :-) – user 1357113 Dec 27 '12 at 18:37

Hints:

You can use "$u$-substitution" to simplify matters:

Let $u = 1 + x^2$. Then $du = \;\;?\;dx,\;\;$ and note that $$\int_0^{2\sqrt{2}} \frac{x}{\sqrt{1+x^2}}dx = \frac{1}{2}\int_0^{2\sqrt{2}} \frac{2x\; dx}{\sqrt{(1+x^2)}}$$

Integrate with respect to $u$, with the new bounds of integration being:

from $\;u\;$ evaluated at $\;x = 0: \; u = 1 + 0^2 = 1\;$
to $\;u\;$ evaluated at $\;x = 2\sqrt{2}:\;1 + (2\sqrt{2})^2 = 1 + 8 = 9$.

There is no need for substitutions $$\int_0^{2\sqrt{2}} \frac{(1+x^2)'}{2\sqrt{1+x^2}}dx=\int_0^{2\sqrt{2}} ({\sqrt{1+x^2}})'dx=\sqrt{1+x^2}|_0^{2\sqrt{2}}=2$$

• I didn't see like this way. ;-) – mrs Dec 27 '12 at 18:32
• @BabakSorouh: this is my favorite way of tackling such integrals :) – user 1357113 Dec 27 '12 at 18:39
• Maybe I should learn it soon. Thanks for sharing it here. I love your challenging questions as well. – mrs Dec 27 '12 at 18:41
• @Babak Sorouh: Thank you for that!!! I've just posted 2 interesting limits maybe you wanna see. They are on main. :) – user 1357113 Dec 27 '12 at 18:42