HW3 SOLUTIONS

SHENGWEN WANG

1. ∫1

x3√x2 − 4

dx

Do substitution x = 2 sec θ, dx = tan θ sec θdθ

=

∫tan θ sec θdθ

8 sec3 θ2 tan θ

=

∫cos2 θ

16dθ

=

∫1 + cos 2θ

32dθ

=θ

32+

sin 2θ

64+ C

=sec−1(x2 )

32+

sin[2 sec−1(x2 )]

64+ C

2. ∫x2

√1− x6dx

=

∫1

3

√1− (x3)2d(x3)

Do substitution u = x3

=

∫1

3

√1− u2du

Do substitution u = sin θ, du = cos θdθ

=

∫1

3cos2 θdθ

=

∫1

6(1 + cos(2θ))dθ

=1

6θ +

1

12sin(2θ)

=1

6sin−1(x3) +

1

12sin[2 sin−1(x3)]

=1

6sin−1(x3) +

1

6x3

√1− x2,

where in the last step we used the double angle formula sin(2θ) = 2 sin θ cos θ and theidentity cos θ =

√1− sin2 θ.

1

2 SHENGWEN WANG

3. It should be decomposed to

3 + x

x(x2 + 2x+ 1)=

3 + x

x(x+ 1)2=A

x+

B

x+ 1+

C

(x+ 1)2

4. It should be decomposed to

2x+ 1

x3 + x2 + x=

2x+ 1

x(x2 + x+ 1)=A

x+

Bx+ C

x2 + x+ 1

5. It should be decomposed to

x5 + 1

x2 + x4=x5 + x3 − x3 + 1

x2(x2 + 1)= x+

−x3 + 1

x2(x2 + 1)= x+

A

x+B

x2+Cx+D

x2 + 1

6. ∫1

1 + 3√xdx

Do substitution u = 3√x, x = u3, dx = 3u2du

=

∫3u2du

1 + u

=

∫3u2 + 3u− 3u− 3 + 3

1 + u

=

∫3udu−

∫3du+

∫3

1 + udu

=3u2

2− 3u+ 3 ln(1 + u) + C

=3x

23

2− 3x

13 + 3 ln(1 + x

13 ) + C

7. ∫1√

x+ 2 + xdx

Do substitution u =√x+ 2, x = u2 − 2, dx = 2udu

=

∫2udu

u+ u2 − 2

=

∫2udu

(u+ 2)(u− 1)

SolveA

u+ 2+

B

u− 1=

2u

(u+ 2)(u− 1)

Get

A =4

3, B =

2

3

HW3 SOLUTIONS 3

So ∫2udu

(u+ 2)(u− 1)

=

∫4

3(u+ 2)du+

∫2

3(u− 1)du

=4

3ln(u+ 2) +

2

3ln(u− 1) + C

=4

3ln(√x+ 2 + 2) +

2

3ln(√x+ 2− 1) + C

8.

∫x2 + 1

x2 − 2x+ 3dx

=

∫(x2 − 2x+ 3) + (2x− 2)

x2 − 2x+ 3dx

=

∫1dx+

∫2x− 2

x2 − 2x+ 3dx

=

∫1dx+

∫2(x− 1)

(x− 1)2 + 2d(x− 1)

=x+ C +

∫d[(x− 1)2]

(x− 1)2 + 2

=x+ ln[(x− 1)2 + 2] + C

9.

∫ 3

1

x3 + 2x2 + x− 1

x3 + xdx

=

∫ 3

1

1dx+

∫ 3

1

2x2 − 1

x3 + xdx

We want to find A,B,C such that

2x2 − 1

x3 + x=

2x2 − 1

x(x2 + 1)=A

x+Bx+ C

x2 + 1

So

A

x+Bx+ C

x2 + 1=Ax2 +A+Bx2 + Cx

x(x2 + 1)

A = −1, B = 3, C = 0

4 SHENGWEN WANG

The integration becomes ∫ 3

1

x3 + 2x2 + x− 1

x3 + xdx

=

∫ 3

1

1dx+

∫ 3

1

−1x

+3x

x2 + 1dx

=

∫ 3

1

1dx+

∫ 3

1

−1xdx+

∫ 3

1

32d(x

2)

x2 + 1

=2− lnx|31 +3

2ln(x2 + 1)|31

=2− ln 3 +3

2ln 10− 3

2ln 2