How do you integrate 1/(x^2 + 10) ?

x^2 + 10 = 10[ (x /√10)^2 + 1 ]

let x /√10 = u

(1 /√10) dx = du

dx = √10 du

∫ dx /(x^2 + 10)

= (1/10)∫ dx /[ (x /√10 )^2 + 1]

=√10/ 10 ∫ du / (u^2 + 1)

= (1 / √10 ) tan^-1(u)

= (1 / √10 ) tan^-1( x /√10 ) + c

There is a formula: ∫dx/(x²+a²)=1/a • arctan(x/a)+C.

Here a=sqrt(10)=√10

If you are US, arctan is written tan^(-1).

By hand (which amounts to proving the formula), for ∫dx/(x²+a²) , substitute x=a•tant, dx=a•dt/cos²(t) to get:

∫dx/(x²+a²) = ∫adt/(a²•cos²(t)•(tan²(t)+1))=1/a • ∫dt = 1/a • t+C.

By the substitution tan(t)=x/a so t=arctan(x/a)

BTW: This is a really cool function. Graph it using GeoGebra - it is upside down bell looking thing that flattens out towards 0 at both ends. The cool thing is that even though the function never stops at both ends, the area between the curve and the x-axis is actually a finite number (the definite integral from -∞ to +∞). Here it is π/√10.