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

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Answer 1 · 2016-12-04T00:38:28

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

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

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

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

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

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

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

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