1. Section 7.5: Review

2. 3𝑥 𝑥 𝑑𝑥
Example 1
Solution a: Substitution
𝑢= 𝑥 2 +4
𝑑𝑥= 𝑑𝑢 2𝑥
𝑑𝑢=2𝑥𝑑𝑥
3𝑥 𝑥 𝑑𝑥= 3𝑥 𝑢 𝑑𝑢 2𝑥
= 𝑢 𝑑𝑢
= 𝑢 − 𝑑𝑢
= 𝑢 − − 𝐶
=3 𝑢 𝐶
=3 𝑥 𝐶

3. Solution b: Trig Substitution
3𝑥 𝑥 𝑑𝑥=
3𝑥 𝑥 𝑑𝑥
𝑥=2 𝑡𝑎𝑛 𝜃
𝑑𝑥=2 𝑠𝑒𝑐 2 𝜃𝑑𝜃
𝑥 = 4 𝑡𝑎𝑛 2 𝜃+4 = 4( 𝑡𝑎𝑛 2 𝜃+1) =2𝑠𝑒𝑐𝜃
3𝑥 𝑥 𝑑𝑥= 3(2𝑡𝑎𝑛𝜃) 2𝑠𝑒𝑐𝜃 2 𝑠𝑒𝑐 2 𝜃𝑑𝜃
=6 𝑡𝑎𝑛𝜃 𝑠𝑒𝑐𝜃 𝑑𝜃
=6𝑠𝑒𝑐𝜃+𝐶
=3 2𝑠𝑒𝑐𝜃 +𝐶
=3 𝑥 𝐶

4. 𝑥 3 ln 𝑥𝑑𝑥 = 𝑥 4 4 ln 𝑥 − ( 𝑥 4 4 )( 1 𝑥 )𝑑𝑥
Example 2
Solution: Integration by parts
𝑢= ln 𝑥 𝑑𝑣= 𝑥 3 𝑑𝑥
𝑑𝑢= 1 𝑥 𝑣= 𝑥 4 4
𝑥 3 ln 𝑥𝑑𝑥 = 𝑥 ln 𝑥 − ( 𝑥 )( 1 𝑥 )𝑑𝑥
= 𝑥 ln 𝑥 − 𝑥 3 𝑑𝑥
= 𝑥 ln 𝑥 − 𝑥 C
= 𝑥 ln 𝑥 − 𝑥 𝐶

5. Num degree > Den degree
Example 3
Solution:
Num degree > Den degree
Apply long division algorithm:
2 𝑥 5 −3𝑥+4 𝑥( 𝑥 = 2 𝑥 5 −3𝑥+4 𝑥 3 +4𝑥
First divide the leading term 2𝑥 5 of the numerator polynomial
by the leading term 𝑥 3 of the divisor and write the answer 2𝑥 2
on the top line:
2𝑥 2
Now multiply this term 2 𝑥 2 by the divisor 𝑥 3 +4𝑥, and write
the answer 2𝑥 2 𝑥 3 +4𝑥 = 2𝑥 5 + 8𝑥 3 under the numerator
polynomial, lining up terms of equal degree:

6. 2𝑥 2
2𝑥 5 + 8𝑥 3
Next subtract the last line from line above it:
2𝑥 2
2𝑥 5 + 8𝑥 3
Now repeat the procedure: divide the leading term −8𝑥 3 of the
polynomial on the last line by the leading term 𝑥 3 of the divisor
to obtain −8 and add this term −8 on the top line:
2𝑥 2 −8
2𝑥 5 + 8𝑥 3

7. Then multiply “back”: −8 𝑥 3 +4𝑥 = −8𝑥 3 −32𝑥 and write the
answer under the last line polynomial, lining up terms of
equal degree:
2𝑥 2 −8
2𝑥 5 + 8𝑥 3
−8𝑥 3 −32𝑥
Next subtract the last line from line above it:
2𝑥 2 −8
2𝑥 5 + 8𝑥 3
−8𝑥 3 −32𝑥
29𝑥+4

8. Num degree < Den degree
You’re done since the degree of 29𝑥+4 is 1, which is less than
the degree of divisor, which is 3.
Consequently,
2 𝑥 5 −3𝑥+4 𝑥( 𝑥 = 2 𝑥 5 −3𝑥+4 𝑥 3 +4𝑥 = 2𝑥 2 −8+ 29𝑥+4 𝑥 3 +4𝑥
2 𝑥 5 −3𝑥+4 𝑥( 𝑥 𝑑𝑥= 2𝑥 2 −8+ 29𝑥+4 𝑥 3 +4𝑥 𝑑𝑥
= 2𝑥 2 𝑑𝑥− 8𝑑𝑥 𝑥+4 𝑥 3 +4𝑥 𝑑𝑥
Next, apply partial fraction for 𝟐𝟗𝒙+𝟒 𝒙 𝟑 +𝟒𝒙
Num degree < Den degree

9. 29𝑥+4 𝑥 3 +4𝑥 = 29𝑥+4 𝑥( 𝑥 2 +4) = 𝐴 𝑥 + 𝐵𝑥+𝐶 𝑥 2 +4
= 𝐴 𝑥 𝐵𝑥+𝐶 𝑥 𝑥( 𝑥 2 +4)
Thus,
𝐴 𝑥 𝐵𝑥+𝐶 𝑥=29𝑥+4
𝐴+𝐵 𝑥 2 +𝐶𝑥+4𝐴=0 𝑥 2 +29𝑥+4
𝐴+𝐵=0
𝐶=29
4𝐴=4
𝐴=1, 𝐵=−1, 𝐶=29
Therefore,
29𝑥+4 𝑥 3 +4𝑥 = 29𝑥+4 𝑥( 𝑥 2 +4) = 𝐴 𝑥 + 𝐵𝑥+𝐶 𝑥 2 +4 = 1 𝑥 + (−1)𝑥+29 𝑥 2 +4
= 1 𝑥 + (−1)𝑥 𝑥 𝑥 2 +4

10. 2 𝑥 5 −3𝑥+4 𝑥( 𝑥 2 +4) 𝑑𝑥= 2𝑥 2 𝑑𝑥− 8𝑑𝑥+ 29𝑥+4 𝑥 3 +4𝑥 𝑑𝑥
= 2𝑥 2 𝑑𝑥− 8𝑑𝑥 𝑥 𝑑𝑥+ (−1)𝑥 𝑥 2 +4 𝑑𝑥 𝑥 2 +4 𝑑𝑥
=2 𝑥 −8𝑥+ ln 𝑥 − ln 𝑥 tan −1 𝑥 2 +𝐶

11. Example 4
Solution a: Known formula 𝑥 𝑥 2 −1 𝑑𝑥= sec −1 𝑥+𝐶
3 𝑥 𝑥 2 −1 𝑑𝑥= 𝑥 𝑥 2 −1 𝑑𝑥
=3 sec −1 𝑥+𝐶
Solution b: Trig Substitution
𝑥=𝑠𝑒𝑐𝜃
𝜃= sec −1 𝑥
𝑑𝑥=𝑠𝑒𝑐𝜃 𝑡𝑎𝑛𝜃 𝑑𝜃
𝑥 2 −1 =𝑡𝑎𝑛𝜃

12. 3 𝑥 𝑥 2 −1 𝑑𝑥= 𝑥 𝑥 2 −1 𝑑𝑥
= 𝑠𝑒𝑐𝜃 𝑡𝑎𝑛𝜃 𝑠𝑒𝑐𝜃 𝑡𝑎𝑛𝜃 𝑑𝜃
=3 1𝑑𝜃
=3𝜃+𝐶
=3 sec −1 𝑥+𝐶

13. Num degree < Den degree
Example 5
Solution: Partial Fraction
Num degree < Den degree
𝑥 2 −4 𝑥( 𝑥 2 +5) 𝑑𝑥= 𝐴 𝑥 + 𝐵𝑥+𝐶 𝑥 2 +5 𝑑𝑥
𝑥 2 −4 𝑥( 𝑥 2 +5) = 𝐴 𝑥 + 𝐵𝑥+𝐶 𝑥 2 +5 = 𝐴 𝑥 𝐵𝑥+𝐶 𝑥 𝑥( 𝑥 2 +5)
1𝑥 2 +0𝑥−4= 𝐴+𝐵 𝑥 2 +𝐶𝑥+5𝐴
1=𝐴+𝐵
0=𝐶
−4=5𝐴
𝐴=− 𝐵= 𝐶=0

14. 𝑥 2 −4 𝑥( 𝑥 2 +5) 𝑑𝑥= 𝐴 𝑥 + 𝐵𝑥+𝐶 𝑥 2 +5 𝑑𝑥
Thus,
𝑥 2 −4 𝑥( 𝑥 2 +5) 𝑑𝑥= 𝐴 𝑥 + 𝐵𝑥+𝐶 𝑥 2 +5 𝑑𝑥
= − 4 5 𝑥 𝑥 𝑥 2 +5 𝑑𝑥
= − 4 5 𝑥 𝑑𝑥 𝑥 𝑥 𝑑𝑥
=− 4 5 ln 𝑥 ln 𝑥 𝐶

15. Solution: Complete the Square
Example 6
Solution: Complete the Square
𝑥 2 +6𝑥−5= 𝑥 2 +6𝑥 − −5
= 𝑥 2 +6𝑥+9−9−5
= (𝑥+3) 2 −14
−4 𝑥 2 +6𝑥−5 𝑑𝑥= −4 (𝑥+3) 2 −14 𝑑𝑥
Then Substitution 𝑢=𝑥+3
𝑑𝑢=𝑑𝑥
−4 𝑥 2 +6𝑥−5 𝑑𝑥= −4 (𝑥+3) 2 −14 𝑑𝑥=− 𝑢 2 − 𝑑𝑢

16. 𝑠𝑒𝑐𝜃= 𝑢 14
Next Trig Substitution 𝑢= 14 𝑠𝑒𝑐𝜃
𝑑𝑢= 14 𝑠𝑒𝑐𝜃 𝑡𝑎𝑛𝜃 𝑑𝜃
𝑢 2 − = 14( 𝑠𝑒𝑐 2 𝜃−1) = 14 𝑡𝑎𝑛𝜃
𝑡𝑎𝑛𝜃= 𝑢 2 −
−4 𝑥 2 +6𝑥−5 𝑑𝑥= −4 (𝑥+3) 2 −14 𝑑𝑥=− 𝑢 2 − 𝑑𝑢
Thus,
=− 𝑡𝑎𝑛𝜃 𝑠𝑒𝑐𝜃 𝑡𝑎𝑛𝜃 𝑑𝜃
=−4 𝑠𝑒𝑐𝜃 𝑑𝜃
=−4 ln 𝑠𝑒𝑐𝜃+𝑡𝑎𝑛𝜃 +𝐶
=−4 ln 𝑢 𝑢 2 − 𝐶
=−4 ln 𝑥 (𝑥+3) 2 − 𝐶