$\int \frac{2x}{x^2+1} \mathrm{d}x$

Project ID: 
3000020119
Question: 

Students Alice, Bob, Christine, and David were tasked with calculating the indefinite integral: $$\int \frac{2x}{x^2+1} \mathrm{d}x$$

They solved it in the following ways:

Alice remembered that if we integrate a function of the form $\frac{g'}{g}$ (a fraction where the numerator is the derivative of the denominator), the integral is equal to $\ln⁡|g|+c$. She then concluded: $$ \int \frac{2x}{x^2+1} \mathrm{d}x=\ln⁡|x^2+1|+c,c\in\mathbb{R}. $$

Bob realized that this is the integral of the quotient of two functions. He integrated the numerator and denominator separately and got: $$ \int \frac{2x}{x^2+1} \mathrm{d}x=\frac{x^2}{\frac{x^3}{3}+x}+c, c\in\mathbb{R}. $$

Christine decided to apply the substitution $x^2+1=t$. She determined the relation between differentials $2x\mathrm{d}x=\mathrm{d}t$ and used the substitution to solve the integral: $$ \int \frac{2x}{x^2+1} \mathrm{d}x=\int \frac{\mathrm{d}t}{t}=\ln⁡|t|+c=\ln⁡|x^2+1|+c,c\in\mathbb{R}. $$

David modified the expression in the integral to the sum of two fractions which he integrated separately: $$ \int \frac{2x}{x^2+1} \mathrm{d}x=\int \left(\frac{2x}{x^2} +\frac{2x}{1}\right) \mathrm{d}x=\int \left(\frac{2}{x}+2x\right)\mathrm{d}x=2 \ln⁡|x|+x^2+c, c\in\mathbb{R}. $$

Who among the students solved the integral correctly? Consider the whole calculation, not just the result.

Answer 1: 

Both, Alice and Christine

Answer 2: 

Only Alice

Answer 3: 

Only Bob

Answer 4: 

Only Christine

Answer 5: 

Only David

Answer 6: 

Nobody

Correct Answer: 
Answer 1
Hint: 

Christine and Alice used the equivalent procedure to solve the integral. Both did the correct calculation and got the right result, valid on the interval $(-\infty,+\infty)$. David and Bob, both made fundamental mistakes. Bob did not realize that the integral of the quotient of two functions is not equal to the quotient of the two corresponding integrals. David performed the incorrect modification. The equality $\frac{2x}{x^2+1}=\frac{2x}{x^2} +\frac{2x}{1}$ is not true in general.