Evaluate the following integrals:


To Find :


Now, let 4x + 1 be written as 2(2x - 1) + 3 and split


Therefore ,


=


=


Now solving,


Let = u dx =


Thus, becomes


Now , = = ) =


=


Now solving,


Now, can be written as


i.e,


Here , let x - = y dx = dy


Therefore, can be written as


Formula Used: = log |x +|+ C


Since is of the form with change in variable.


= log |y +|+ C


= log |y +|+ C


Since , x - = y and dx = dy


= log |(x - ) +|+C


Therefore,


= log |x - +|+C


Hence ,


= + log |x - +|+C


Therefore ,


= + log |x - +|+C


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