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

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1 Answer
Jun 27, 2018

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

Explanation:

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