Summary MAT2691 ASSIGNMENT 2 2024 SOLUTIONS

MAT2691
ASSIGNMENT 2
2024

, QUESTION 1




Solution:



1).



x3
∫ dx
1 + x4


du 1
Let: u = 1 + x 4 ⇒ = 4x 3 ⇒ dx = du
dx 4x 3

, x3 x3 1
∫ dx = ∫ ∙ du
1 + x4 u 4x 3
1 1
= ∫ du
4 u
1
= ln|u| + C
4
1
= ln|1 + x 4 | + C
4


2).



x4
∫ dx
1 + x3


x4 x
3
=x−
1+x 1 + x3


x4 x
∫ 3
dx = ∫ [x − ] dx
1+x 1 + x3
x
= ∫ x dx − ∫ dx
1 + x3
x2 x
= −∫ dx
2 1 + x3


x
Decompose into partial fractions:
1 + x3


x x
3
=
1+x (x + 1)(x 2 − x + 1)
A Bx + C
= + 2
(x + 1) (x − x + 1)
A(x 2 − x + 1) + (Bx + C)(x + 1)
=
(x + 1)(x 2 − x + 1)
Ax 2 − Ax + A + Bx 2 + Bx + Cx + C
=
(x + 1)(x 2 − x + 1)
(A + B)x 2 + (−A + B + C)x + (A + C)
=
(x + 1)(x 2 − x + 1)

, x (A + B)x 2 + (−A + B + C)x + (A + C)
=
(x + 1)(x 2 − x + 1) (x + 1)(x 2 − x + 1)



By comparing both sides of the equation:



A+B =0 1

−A + B + C = 1 2

A+C=0 3



From 1 B = −A and from 3 C = −A then substitute B and C into 2



−A + B + C = 1 2

−A + (−A) + (−A) = 1
−3A = 1
1
A=−
3




B = −A
1
B = − (− )
3
1
B=
3


C = −A
1
C = − (− )
3
1
C=
3


x x
3
=
1+x (x + 1)(x 2 − x + 1)
x A Bx + C
3
= + 2
1+x (x + 1) (x − x + 1)