At first glance, this may look like an inverse trig integral but as the top contains an x term, we must use a different method.
First, rewrite the top in two parts, one that is a multiple of the derivative of the bottom and one that is just a number. In this case the derivative of the bottom is 2x+2 so rewrite as (2x+2)/2+3.
Now split the fraction in two: (2x+2)/2(x2+2x+2)+3/(x2+2x+2). The first part is now a log integral integrating to 1/2 ln(x2+2x+2).
To integrate the second we must complete the square on the bottom. x2+2x+2=(x+1)2+1. Now using the substitution u=x+1, dx=du, we can see that the second part is an inverse tan integral integrating to arctan(x+1).
The overall integral is therefore 1/2 ln(x2+2x+2) + arctan(x+1).
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