Can someone please help me out this question last attempt.?

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Answer 1 · 2018-06-27T01:44:58

#int(1/(16u)+9/(16u^2))du#

#int((x+3)/(4x+3)^2)dx#

Let #u=4x+3# therefore #(u-3)/4=x# and #(du)/4=dx#

#1/4int(((u-3)/4+3)/u^2)du#

#1/4int(((u-3)/4+12/4)/u^2)du#

#1/4int(((u+9))/(4u^2))du#

#1/16int(1/u+9/u^2)du#

Bring #1/16# back in to fit your box

#int(1/(16u)+9/(16u^2))du#