Advanced book on Mathematics Olympiad

532 Real Analysis

444.Split the integral as
∫
ex
2
dx+

∫

2 x^2 ex

2

dx.

Denote the first integral byI 1. Then use integration by parts to transform the second
integral as
∫
2 x^2 ex
2
dx=xex
2
−

∫

ex

2

dx=xex

2

−I 1.

The integral from the statement is therefore equal to

I 1 +xex

2

−I 1 =xex

2

+C.

445.Adding and subtractingexin the numerator, we obtain
∫
x+sinx−cosx− 1
x+ex+sinx

dx=

∫

x+ex+sinx− 1 −ex−cosx

x+ex+sinx

dx

=

∫

x+ex+sinx

x+ex+sinx

dx−

∫

1 +ex+cosx

x+ex+sinx

dx

=x+ln(x+ex+sinx)+C.

(Romanian college entrance exam)

446.The trick is to bring a factor ofxinside the cube root:
∫
(x^6 +x^3 )^3

√

x^3 + 2 dx=

∫

(x^5 +x^2 )^3

√

x^6 + 2 x^3 dx.

The substitutionu=x^6 + 2 x^3 now yields the answer

1

6

(x^6 + 2 x^3 )^4 /^3 +C.

(G.T. Gilbert, M.I. Krusemeyer, L.C. Larson,The Wohascum County Problem Book,
MAA, 1993)

447.We want to avoid the lengthy method of partial fraction decomposition. To this end,
we rewrite the integral as

∫ x^2

(

1 +

1

x^2

)

x^2

(

x^2 − 1 +

1

x^2

)dx=

∫ 1 +^1

x^2

x^2 − 1 +

1

x^2

dx.