2013-01-26T10:20:27-05:00
Find the integral of ∫(y-5) dy ∕ y^2 + 5y + 6
1
2013-01-29T11:17:52-0500
∫ y − 5 y 2 + 5 y + 6 d y = … \int \frac {y - 5}{y ^ {2} + 5 y + 6} d y = \dots ∫ y 2 + 5 y + 6 y − 5 d y = … y − 5 y 2 + 5 y + 6 = y − 5 ( y + 2 ) ( y + 3 ) = A y + 2 + B y + 3 = ( A + B ) y + ( 3 A + 2 B ) ( y + 2 ) ( y + 3 ) \frac {y - 5}{y ^ {2} + 5 y + 6} = \frac {y - 5}{(y + 2) (y + 3)} = \frac {A}{y + 2} + \frac {B}{y + 3} = \frac {(A + B) y + (3 A + 2 B)}{(y + 2) (y + 3)} y 2 + 5 y + 6 y − 5 = ( y + 2 ) ( y + 3 ) y − 5 = y + 2 A + y + 3 B = ( y + 2 ) ( y + 3 ) ( A + B ) y + ( 3 A + 2 B ) { A + B = 1 3 A + 2 B = − 5 \left\{ \begin{array}{l} A + B = 1 \\ 3 A + 2 B = - 5 \end{array} \right. { A + B = 1 3 A + 2 B = − 5 { 2 A + 2 B = 2 3 A + 2 B = − 5 \left\{ \begin{array}{l} 2 A + 2 B = 2 \\ 3 A + 2 B = - 5 \end{array} \right. { 2 A + 2 B = 2 3 A + 2 B = − 5 − A = 7 - A = 7 − A = 7 A = − 7 , B = 8 A = - 7, B = 8 A = − 7 , B = 8 y − 5 y 2 + 5 y + 6 = − 7 y + 2 + 8 y + 3 \frac {y - 5}{y ^ {2} + 5 y + 6} = \frac {- 7}{y + 2} + \frac {8}{y + 3} y 2 + 5 y + 6 y − 5 = y + 2 − 7 + y + 3 8 … = ∫ ( − 7 y + 2 + 8 y + 3 ) d y = − 7 ln ∣ y + 2 ∣ + 8 ln ∣ y + 3 ∣ + C , C = c o n s t \ldots = \int \left(\frac {- 7}{y + 2} + \frac {8}{y + 3}\right) d y = - 7 \ln | y + 2 | + 8 \ln | y + 3 | + C, C = c o n s t … = ∫ ( y + 2 − 7 + y + 3 8 ) d y = − 7 ln ∣ y + 2∣ + 8 ln ∣ y + 3∣ + C , C = co n s t
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