How do you integrate #int_1^7 sqrtx*lnx # using integration by parts?

1 Answer
Andrea S.
Dec 4, 2016

#int_1^7sqrt(x)lnxdx = 2/3sqrt(343)(ln7 - 2/3)+4/9#

Explanation:

#int_1^7sqrt(x)lnxdx = 2/3int_1^7lnx d(x^(3/2))#

#int_1^7sqrt(x)lnxdx = 2/3(x^(3/2)lnx )|_(x=1)^(x=7) - 2/3int_1^7x^(3/2)d(lnx)#

#int_1^7sqrt(x)lnxdx = 2/3(x^(3/2)lnx )|_(x=1)^(x=7) - 2/3int_1^7x^(3/2)dx/x#

#int_1^7sqrt(x)lnxdx = 2/3(x^(3/2)lnx )|_(x=1)^(x=7) - 2/3int_1^7x^(1/2)dx#

#int_1^7sqrt(x)lnxdx = 2/3x^(3/2)(lnx - 2/3)|_(x=1)^(x=7)#

#int_1^7sqrt(x)lnxdx = 2/3sqrt(343)(ln7 - 2/3)+4/9#

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