Advanced book on Mathematics Olympiad

Real Analysis 533

With the substitutionx−^1 x=twe have( 1 +x^12 )dx=dt, and the integral takes the form

∫

1

t^2 + 1

dt=arctant+C.

We deduce that the integral from the statement is equal to

arctan

(

x−

1

x

)

+C.

448.Substituteu=

√

ex− 1

ex+ 1 ,0<u<1. Thenx =ln(^1 +u

(^2) )−ln( 1 −u (^2) ), and
dx=( 1 +^2 uu 2 + 1 −^2 uu 2 )du. The integral becomes
∫
u

(

2 u

u^2 + 1

+

2 u

u^2 − 1

)

du=

∫ (

4 −

2

u^2 + 1

+

2

u^2 − 1

)

du

= 4 u−2 arctanu+

∫ (

1

u+ 1

+

1

1 −u

)

du

= 4 u−2 arctanu+ln(u+ 1 )−ln(u− 1 )+C.

In terms ofx, this is equal to

4

√

ex− 1

ex+ 1

−2 arctan

√

ex− 1

ex+ 1

+ln

(√

ex− 1

ex+ 1

+ 1

)

−ln

(√

ex− 1

ex+ 1

− 1

)

+C.

449.If we naively try the substitutiont=x^3 +1, we obtain

f(t)=

√

t+ 1 − 2

√

t+

√

t+ 9 − 6

√

t.

Now we recognize the perfect squares, and we realize that

f(x)=

√

(

√

x^3 + 1 − 1 )^2 +

√

(

√

x^3 + 1 − 3 )^2 =|

√

x^3 + 1 − 1 |+|

√

x^3 + 1 − 3 |.

Whenx∈[ 0 , 2 ],1≤

√

x^3 + 1 ≤3. Therefore,

f(x)=

√

x^3 + 1 − 1 + 3 −

√

x^3 + 1 = 2.

The antiderivatives offare therefore the linear functionsf(x)= 2 x+C, whereCis a
constant.
(communicated by E. Craina)

450.Letfn= 1 +x+x

2

2 !+···+

xn
n!. Thenf
′(x)= 1 +x+···+xn−^1
(n− 1 )!. The integral
in the statement becomes