Letting t = x^2 + 1 for x/(x^2 + 1) dx gives
Aln(x^2 + 1) + C
B(1/2) tan^-1 x + C
C(1/2) ln(x^2 + 1) + C
D2 ln(x^2 + 1) + C
Answer & Solution
Correct answer: C. (1/2) ln(x^2 + 1) + C
Putting t = x^2 + 1 gives x dx = dt/2, so the integral is (1/2) ln t.
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