Please help with integration problem- driving me crazy!?

Question

Evaluate the indefinate integral of
(1+x)/(1+x^2) dx

The answer is 1/2 ln (1+x^2) + arctan (x)....

Answer 1

write it as
1/(1+x^2)+ x/(1+x^2)

try to differentiate 1/2 ln (1+x^2), what do you notice?

Answer 2

Split the problem into 1/(1+X^2) + X/(1+X^2)

The integral of 1/(1+X^2) is arctan (x), so you are left with
x/(1+x^2). Since the degree of the fraction is x^-1, we know the integral will be in terms of ln (u). So you take the bottom of the problem and put that in the ln(). You then get ln(1+x^2). The 1/2 comes from the derivative of 1+x^2, which is supposed to be the numerator of the fraction. Since the numerator is x, instead of 2x (the derivative of 1+x^2), you have to divide the whole thing by 2, thus giving you your answer.

Answer 3

First, break this into two integrals:

∫x/(x²+1) dx + ∫1/(x²+1) dx

The first integral can be solved with a u-substitution - let u=x²+1, then du=2x dx, so we have:

1/2 ∫1/u du + ∫1/(x²+1) dx

Integrating the first one:

1/2 ln |u| + ∫1/(x²+1) dx
1/2 ln (x²+1) + ∫1/(x²+1) dx

And it pays to memorize the fact that the second one is the derivative of arctan x, but doing a trig sub just to show you why that is, let tan θ = x, so dx = sec² θ dθ, thus we have:

1/2 ln (x²+1) + ∫sec² θ/(tan² θ+1) dθ
1/2 ln (x²+1) + ∫sec² θ/sec² θ dθ
1/2 ln (x²+1) + ∫1dθ
1/2 ln (x²+1) + θ + C
1/2 ln (x²+1) + arctan x + C

And we are done.