Advanced book on Mathematics Olympiad

534 Real Analysis

In=

∫

n!(fn(x)−fn′(x))

fn(x)

dx=n!

∫ (

1 −

fn′(x)

fn(x)

)

dx=n!x−n!lnfn(x)+C

=n!x−n!ln

(

1 +x+

x^2

2!

+···+

xn

n!

)

+C.

(C. Mortici,Probleme Pregatitoare pentru Concursurile de Matematic ̆ ̆a(Training
Problems for Mathematics Contests), GIL, 1999)

451.The substitution is

u=

x

√ (^42) x (^2) − 1 ,
for which
du=
x^2 − 1
( 2 x^2 − 1 )^4

√

2 x^2 − 1

dx.

We can transform the integral as follows:
∫
2 x^2 − 1
−(x^2 − 1 )^2

·

x^2 − 1

( 2 x^2 − 1 )^4

√

2 x^2 − 1

dx=

∫

1

−x^4 + 2 x^2 − 1

2 x^2 − 1

·

x^2 − 1

( 2 x^2 − 1 )^4

√

2 x^2 − 1

dx

=

∫

1

1 − x

4

2 x^2 − 1

·

x^2 − 1

( 2 x^2 − 1 )^4

√

2 x^2 − 1

dx

=

∫

1

1 −u^4

du.

This is computed using Jacobi’s method for rational functions, giving the final answer to
the problem

1

4

ln

√ (^42) x (^2) − 1 +x
√ (^42) x (^2) − 1 −x−

1

2

arctan

√ (^42) x (^2) − 1
x

+C.

452.Of course, Jacobi’s partial fraction decomposition method can be applied, but it is
more laborious. However, in the process of applying it we factor the denominator as
x^6 + 1 =(x^2 + 1 )(x^4 −x^2 + 1 ), and this expression can be related somehow to the
numerator. Indeed, if we add and subtract anx^2 in the numerator, we obtain

x^4 + 1

x^6 + 1

=

x^4 −x^2 + 1

x^6 + 1

+

x^2

x^6 + 1

.

Now integrate as follows:
∫
x^4 + 1
x^6 + 1

dx=

∫

x^4 −x^2 + 1

x^6 + 1

dx+

∫

x^2

x^6 + 1

dx=

∫

1

x^2 + 1

dx+

∫

1

3

(x^3 )′

(x^3 )^2 + 1

dx