∫ (1+x)/(1+x^2), u substitution?

Question

Similar to a previous problem I posted, I rewrote this problem and expressed it as thus:

[ ∫ 1/(1+x^2) + ∫ x/(1+x^2) ] dx

Then I remember that ∫ 1/(1+x^2) = arctan x

and so I came to

arctanx + [(1/2)∫ 1/(1+x^2)]

and I get

arctanx + (1/2)arctanx?

But the book is arctanx + (1/2) ln | 1+x^2 | + C

Again, why, in this case, would it be (1/2) ln | 1+x^2 | rather than (1/2)arctanx?

I forgot to include a step.

I didn't include the x in the second integral because when I used U substitution

u = 1+ x^2 and du = 2xdx

thus ∫ x/u du/2x

Doesn't the x cancel out to be

(1/2) ∫ 1/u du ?

Then.. that would be (1/2) lnu

Oh... nevermind!

Answer 1

The book answer is correct,

Let's take your 2nd integral,
∫ x/(1 + x^2) dx

After u substitute, u = 1 + x^2 and du/2 = x dx, your 2nd integral becomes,

1/2 ∫ du/u
= 1/2(ln(u))
= 1/2 ln(1 + x^2)

Answer 2

You have two separate integrals. The first is arctan(x)
now the second, integ x/1+x^2 , let x^2=u
2xdx=du
xdx=(1/2)du
=1/2 integ du/1+u =1/2 ln|1+u|=1/2ln|1+x^2|
So the book is OK

Answer 3

You forgot the x in your second integral!
∫ x/(1+x^2) dx =1/2 ∫ 2x/(1+x^2) ] dx.
Now notice that 2x is the derivative of 1+x².
So your second integral is of the form ∫ du/u.
So the final result is 1/2 ln(1+x²) + C.
Here you don't even need the absolute value
signs, because 1+x² is always nonnegative.

Answer 4

You have the right idea.

∫ (1+x)/(1+x^2) dx

= ∫ (1)/(1+x^2) dx + ∫ (x)/(1+x^2) dx

= arctan(x) + C + ∫ (x)/(1+x^2) dx
Let u = 1 + x^2
du = 2x dx

= arctan(x) + C + (1/2) ∫ (1)/(u) du
= arctan(x) + C + (1/2) ln(u) + C
= arctan(x) + C + (1/2) ln(1 + x^2) + C
= arctan(x) + (1/2) ln(1 + x^2) + C

Answer 5

You dropped the x in the numerator of the second integral. That's why it's not arctanx.

Also, I'm not following where the (1/2) came from in front of the second integral.