Evaluate the following integral for 
:
![\[ \int \frac{dx}{a^2 + x^2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ddb576a602b660a7fa2d8cddbfe91f9e_l3.png "Rendered by QuickLaTeX.com")
Since we may write
![\[ \int \frac{dx}{a^2+x^2} = \int \frac{1}{1+\left(\frac{x}{a}\right)^2} \frac{dx}{a^2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-89ef176529599265933e4053483c061b_l3.png "Rendered by QuickLaTeX.com")
Now, make the substitution 
and 
. Then we have,
Evaluate the following integral for :
Since we may write
Now, make the substitution and . Then we have,
Why did 1/a^2 become 1/a ??
bcz when u=x/a du/dx=1/a du=dx/a this a is used from 1/a^2
You have to divide the whole expression by 1 / a since it is (x/a)^2.
Thank u it helped me
Good work
arc means???
arctan is inverse tan
Yup dude arctan is tan-¹(x/y)