Help with this integral?

Question

How do I go about solving these two integrals:

1. THe integral from 2x -> 3x OF (u^2 -1) / (u^2 + 1)

2. Suppose the graph of a smooth curve y = f(x) passes through the point (0,1). Let a and b be such that f(a) = 6 and the integral from 0 -> a of f'(x)dx = 3 times the integral of 0 -> b of f'(x)dx. Find F(b)

I understand how the following people are answering the question except for one part.

In the first question how they get the line:
u - 2∫du/(u²+1)

Answer 1

∫(u²-1)/(u²+1)du
=∫(u²+1-2)/(u²+1)du
= u - 2∫du/(u²+1)
= u - 2 arctan(u) from 2x to 3x
= x - 2 arctan(3x) + 2 arctan(2x) + C

[0 to a]∫f'(x)dx = [0 to b]∫3f'(x)dx
f(a) - f(0) = 3(f(b) - f(0))
6 - 1 = 3(f(b)) - 3
8 = 3f(b)
f(b) = 8/3

Answer 2

1. Let's do the indefinite integral, then plug in the limits.
We must divide u²+ 1 into u²-1 to reduce the degree.
So, by long division, we get

.............1.............................
u² + 1 | u² - 1
..........| u² + 1
.............................
............ ...... -2

So the result is 1 - 2/(u²+1).
An antiderivative of that is u -2 arctan(u)
and your integral evaluates to
3x -2 arctan 3x -2x + 2 arctan 2x =
x -2(arctan 3x -arctan 2x).

Question 2:
I will assume you are asking for f(b).
First, f(0) = 1, and, by the fundamental theorem
of calculus, we get
f(a)-f(0) = 3(f(b)-f(0) ).
So
6-1 = 3f(b) -3
3f(b) = 8
f(b) = 8/3.
Hope that helps!

Answer 3

(u^2-1)/^(u^+1)= 1-2(1/(1+u^2)
Int = u-2Arctan u (between 3x and 2x)=
=x-2(Arctan2x-Arctan3x)
2) Int(0 =>a)f´(x) = f(a)-f(0)
Int(0=>b) f´(x) = f(b)-f(0)
f(a)-f(0) = 3[f(b)-f(0)]
f(0)=1 and f(a)=6
so 5 =3 f(b)-3 so f(b) =8/3
to1) as it is the definite integral NO constant added