How do you find the definite integral of #intx^3 sqrt(x^2 + 1 )dx# from #[0,sqrt3]#?

1 Answer
Ananda Dasgupta
Mar 6, 2018

#58/15#

Explanation:

Substitute #x^2+1 = u^2#. Then #2xdx = 2udu# and the limits #x=0# and #x = sqrt(3)# becomes #u=1# and #u=2#, respectively.

Thus

# int_0^{sqrt(3)}x^3 sqrt(x^2 + 1 )dx = int_0^{sqrt(3)}x^2 sqrt(x^2 + 1 )quad xdx #
# qquad = quad int_1^2 (u^2-1)u times udu = int_1^2 u^4 du - int_1^2 u^2du#
# qquad = (u^5/5-u^3/3)_1^2 = (32/5-8/3)-(1/5-1/3)#
# qquad = 56/15-(2/15) = 58/15#

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