integration of x2+1


The integral

I:=∫x2+1⁒𝑑x

can be found by using the first Euler’s substitution (http://planetmath.org/EulersSubstitutionsForIntegration)

x2+1:=-x+t,

but another possibility is to use partial integration (http://planetmath.org/ASpecialCaseOfPartialIntegration) if one knows the integral ∫d⁒xx2+1.  The corresponding may be said of the more general

∫x2+c⁒𝑑x.

We think that the integrand of I has the other factor 1 and integrate partially:

I=∫1β‹…x2+1⁒𝑑x=x⁒x2+1-∫xβ‹…12⁒x2+1β‹…2⁒x⁒𝑑x+Cβ€²=x⁒x2+1-∫x2x2+1⁒𝑑x+Cβ€².

Writing the numerator as (x2+1)-1 and dividing its minuend and subtrahend separately, we can write

I=x⁒x2+1-(∫x2+1⁒𝑑x-∫1x2+1⁒𝑑x)+Cβ€²=x⁒x2+1-I+∫d⁒xx2+1+Cβ€².

Having I in two , we solve it from these equalities, obtaining

I=x2⁒x2+1+12⁒∫d⁒xx2+1+C,

i.e.,

∫x2+1⁒𝑑x=x2⁒x2+1+12⁒arsinh⁑x+C
Title integration of x2+1
Canonical name IntegrationOfsqrtx21
Date of creation 2013-03-22 18:06:58
Last modified on 2013-03-22 18:06:58
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 7
Author pahio (2872)
Entry type Derivation
Classification msc 26A36
Synonym antiderivative of x2+1
Related topic IntegrationByParts
Related topic AreaFunctions
Related topic DerivativeOfInverseFunction