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Question
Evaluate the following integral:
`int(4x + 3)/(2x + 1).dx`
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Solution 1
`int(4x + 3)/(2x + 1).dx`
= `int((2(2x + 1) + 1))/(2x + 1).dx`
= `int ((2(2x + 1))/(2x + 1) + 1/(2x + 1)).dx`
= `2 int 1 dx + int 1/(2x + 1).dx`
= `2x + (1)/(2) log|2x + 1| + c`.
Solution 2
`int(4x + 3)/(2x + 1).dx`
`u = 2x + 1=> (du)/(dx) = 2 => dx = (du)/2`
Now express the numerator 4x + 3 in terms of u:
`x = (u-1)/2`
`4x+3=4xx (u-1)/2 +3 = 2(u-1)+3=2u-2+3=2u+1`
`int(4x+3)/(2x+1) dx = int(2u+1)/uxx(du)/2`
`= 1/2int(2+1/u)du`
`1/2 int (2+1/u) du=1/2(2u+ln|u|)+C=u+1/2 ln|u|+C`
`=(2x+1)+1/2ln |2x+1|+C`
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