Section 7.5: Review

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

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

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 4 4 ln 𝑥 − ( 𝑥 4 4 )( 1 𝑥 )𝑑𝑥 = 𝑥 4 4 ln 𝑥 − 1 4 𝑥 3 𝑑𝑥 = 𝑥 4 4 ln 𝑥 − 1 4 𝑥 4 4 +C = 𝑥 4 4 ln 𝑥 − 𝑥 4 16 +𝐶

Num degree > Den degree Example 3 Solution: Num degree > Den degree Apply long division algorithm: 2 𝑥 5 −3𝑥+4 𝑥( 𝑥 2 +4 = 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:

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

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

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 +4 = 2 𝑥 5 −3𝑥+4 𝑥 3 +4𝑥 = 2𝑥 2 −8+ 29𝑥+4 𝑥 3 +4𝑥 2 𝑥 5 −3𝑥+4 𝑥( 𝑥 2 +4 𝑑𝑥= 2𝑥 2 −8+ 29𝑥+4 𝑥 3 +4𝑥 𝑑𝑥 = 2𝑥 2 𝑑𝑥− 8𝑑𝑥+ 29𝑥+4 𝑥 3 +4𝑥 𝑑𝑥 Next, apply partial fraction for 𝟐𝟗𝒙+𝟒 𝒙 𝟑 +𝟒𝒙 Num degree < Den degree

29𝑥+4 𝑥 3 +4𝑥 = 29𝑥+4 𝑥( 𝑥 2 +4) = 𝐴 𝑥 + 𝐵𝑥+𝐶 𝑥 2 +4 = 𝐴 𝑥 2 +4 + 𝐵𝑥+𝐶 𝑥 𝑥( 𝑥 2 +4) Thus, 𝐴 𝑥 2 +4 + 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 2 +4 + 29 𝑥 2 +4

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

Example 4 Solution a: Known formula 1 𝑥 𝑥 2 −1 𝑑𝑥= sec −1 𝑥+𝐶 3 𝑥 𝑥 2 −1 𝑑𝑥=3 1 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

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

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 2 +5 + 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=− 4 5 𝐵= 9 5 𝐶=0

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 + 9 5 𝑥 𝑥 2 +5 𝑑𝑥 = − 4 5 𝑥 𝑑𝑥 + 9 5 𝑥 𝑥 2 +5 𝑑𝑥 =− 4 5 ln 𝑥 + 9 5 1 2 ln 𝑥 2 +5 +𝐶

Solution: Complete the Square Example 6 Solution: Complete the Square 𝑥 2 +6𝑥−5= 𝑥 2 +6𝑥+ 6 2 2 − 6 2 2 −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=−4 1 𝑢 2 − 14 2 𝑑𝑢

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