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Question 1 of 5
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Apply the Addition of Fractions Rule
∫ x3+4x3dx |
= |
∫(x3x3+4x3)dx |
= |
∫(1+4x3)dx |
Use the Sum or Difference Rule
∫(1+4x3)dx |
= |
∫ 1dx+∫ 4x3dx |
Find the Indefinite Integral
∫ 1dx+∫ 4x3dx |
= |
∫x0dx+∫4x−3dx |
Take the constant 4 out of the integral sign |
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= |
∫x0dx+4∫x−3dx |
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= |
x0+10+1+4x−3+1−3+1+c |
Apply the Integration Formula |
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= |
x11+4x-2-2+c |
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= |
x+4(−12)x−2+c |
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= |
x-2x-2+c |
Simplify |
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= |
x-2x2+c |
Apply Negative Indice law |
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Question 2 of 5
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∫ 3(6-x)4dx can be written as ∫ 3(6-x)-4dx
Find the Indefinite Integral
∫3(6−x)−4dx |
= |
3∫(6−x)−4dx |
Take the constant 3 out of the integral sign |
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= |
3(6−x)−4+1(−4+1)(6−x)′+c |
Apply the Integration Formula |
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= |
3(6−x)−3(−3)(−1)+c |
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= |
(6−x)−3+c |
Simplify |
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= |
1(6-x)3+c |
Apply Negative Indice law |
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Question 3 of 5
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Remove the fraction by expressing the term using negative exponents
Find the Indefinite Integral
∫x−4dx |
= |
x−4+1−4+1+c |
Apply the Integration Formula |
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= |
x-3-3+c |
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= |
-13x3+c |
Apply Negative Indice law |
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Question 4 of 5
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∫ 23√xdx can be written as ∫ 23x-12dx
Find the Indefinite Integral
∫23x−12dx |
= |
23∫x−12dx |
Take the constant 23 out of the integral sign |
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= |
23x−12+1(−12+1)+c |
Apply the Integration Formula |
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= |
23x1212+c |
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= |
23(2)x12+c |
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= |
43x12+c |
Simplify |
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= |
43√x+c |
Convert into surd form |
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Question 5 of 5
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∫3√2x-3dx can be written as ∫ 3(2x-3)-12dx
Find the Indefinite Integral
∫3(2x−3)−12dx |
= |
3∫(2x−3)−12dx |
Take the constant 3 out of the integral sign |
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= |
3(2x−3)−12+1(−12+1)(2x−3)′+c |
Apply the Integration Formula |
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= |
3(2x−3)1212×2+c |
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= |
3(2x-3)12+c |
Simplify |
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= |
3√2x-3+c |
Convert into surd form |
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