Question

integration of 1/√1+x^2​

Answer 1

Step-by-step explanation:

The integral of 1 over the square root of 1−x2 is of the form

∫11–x2−−−−√dx=sin–1x+c

To prove this formula, consider

ddx[sin–1x+c]=ddxsin–1x+ddxc

Using the derivative formula ddxsin–1x=11–x2√, we have

ddx[sin–1x+c]=11–x2−−−−√+0⇒ddx[sin–1x+c]=11–x2−−−−√⇒11–x2−−−−√=ddx[sin–1x+c]⇒11–x2−−−−√dx=d[sin–1x+c] – – – (i)

Integrating both sides of equation (i) with respect to x, we have

∫11–x2−−−−√dx=∫d[sin–1x+c]

As we know that by definition the integration is the inverse process of the derivative, so we have

∫11–x2−−−−√dx=sin–1x+c

We know that the derivative of

ddxcos–1x=–11–x2−−−−√

This formula can also be written as

∫11–x2−−−−√dx=–cos–1x+c

Read more: https://www.emathzone.com/tutorials/calculus/integral-of-1-over-square-root-of-1-x2.html#ixzz6nFgn8ejg

Answer 2

Answer:

plz explan step by step