Advanced book on Mathematics Olympiad

Real Analysis 535

=arctanx+

1

3

arctanx^3.

To write the answer in the required form we should have

3 arctanx+arctanx^3 =arctan

P(x)

Q(x)

.

Applying the tangent function to both sides, we deduce

3 x−x^3

1 − 3 x^2 +x

3

1 −^31 −x− 3 xx 23 ·x^3

=tan

(

arctan

P(x)

Q(x)

)

.

From here

arctan

P(x)

Q(x)

=arctan

3 x− 3 x^5

1 − 3 x^2 − 3 x^4 +x^6

,

and henceP(x)= 3 x− 3 x^5 ,Q(x)= 1 − 3 x^2 − 3 x^4 +x^6. The final answer is

1

3

arctan

3 x− 3 x^5

1 − 3 x^2 − 3 x^4 +x^6

+C.

453.The functionf:[− 1 , 1 ]→R,

f(x)=

√ (^3) x
√ (^31) −x+√ (^31) +x,
is odd; therefore, the integral is zero.
454.We use the example from the introduction for the particular functionf(x)= 1 +xx 2
to transform the integral into
π
∫ π 2
0
sinx
1 +sin^2 x
dx.
This is the same as
π
∫ π 2
0

−

d(cosx)

2 −cos^2 x

,

which with the substitutiont=cosxbecomes

π

∫ 1

0

1

2 −t^2

dt=

π

2

√

2

ln

√

2 +t

√

2 −t

∣

∣∣

∣∣

1

0

=

π

2

√

2

ln

√

2 + 1

√

2 − 1

.