I have been doing this question and I am lost as there is still something else that need to do to get the final answer,can you help me. EvaluateAnswer:sdtt² + t-30 Question 8) Evaluatedt1124² ++-30(²+²³² +²6+²-5€-30 +08-7727(((+6)-5(4+6)(6-5) (€ + 6) = 2 lineardegreeA (t=5) (t+6) + B4-5S16+6£√+-s-S11G In t-sl+1Set dt€ ² + € -30TT$AS-7| 1 = A (-6+6) +B(-6-5) → 1 = A (0) + B (11)1 == B|1=A[5+6) + BCS-5) = 1 = A (11) + B (0) Ina-lnb = Ina1α= 49+9+B-9+7LCD= (t =5) (£+6)(t-5) (£+5) = A (t+6) +B(t~5)€+6=0 €-5₂015-7 9=175-7011In 1 € -61 + C →→ !r9-7/11deD

Answered: Evaluate Answer: dt t² + t-30 S₁

Math

Calculus

Evaluate Answer: dt t² + t-30 S₁

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

I have been doing this question and I am lost as there is still something else that need to do to get the final answer,

can you help me.

Question 8) Evaluate
dt
1124² ++-30
(²+²³² +²6+²-5€-30 +
08-7727
(((+6)-5(4+6)
(6-5) (€ + 6) = 2 linear
degree
A (t=5) (t+6) + B
4-5
S
1
6+6
£
√+-s
-S
11
G In t-sl
+
1
Set dt
€ ² + € -30
TT
$
A
S-7
| 1 = A (-6+6) +B(-6-5) → 1 = A (0) + B (11)
1 =
= B
|1=A[5+6) + BCS-5) = 1 = A (11) + B (0) Ina-lnb = Ina
1α= 4
9+9
+
B
-
9+7
LCD= (t =5) (£+6)
(t-5) (£+5) = A (t+6) +B(t~5)
€+6=0 €-5₂0
1
5-7 9=17
5-7011
In 1 € -61 + C →→ !
r
9-7/11
de
D

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$\int \frac{1}{t^{2} + t - 30} d t$

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