Category 2: Substitution 1. Integral: ??2x(x^2+4)^2 ? dx Let u = x^2+4, then du

Question

Category 2:

Substitution

1. Integral: ??2x(x^2+4)^2 ? dx Let u = x^2+4, then du = 2x dx

Substitution: ??u^2 du?

Solution: F(u)=1/3 u^3,

thus F(x)=1/3 (x^2+4)^3+C

2.Integral: ??x?(?3x?^2-5)?dx Let u =?3x?^2-5, then du = 6x dx

Substitution: 1/6 ??6x?(3x^2-5) dx=1/6 ???u du??

Solution: F(u)=1/6[2/3 u^(3/2)],

thus F(x)=1/9 (3x^2-5)^(3/2)+C

3. Integral: ??x^3/(5x^4+9)^5 dx? Let u = 5x^4+9, then du = 20x^3 dx

Substitution: 1/20 ??(20x^3)/(5x^4+9)^5 dx=1/20 ?du/u^5 ?

Solution: F(u)=1/20[-1/(4u^4 )],

thus F(x)=-1/(80?(5x^4+9)?^4 )+ C

How is u determined? How is du determined? How does knowing these two items allow us to make the subsequent substitution? Why did 1/6 and 1/20 appear outside the integration symbol in examples 2 and 3? What condition is necessary to make substitution work? What are the steps in the process of using substitution to integrate? i

Explanation / Answer

in the second problem

he made the substitution u = (3x^2 - 5 )

upon differentiation gives du = 6x dx

so now he has substituted (3x^2 -5) as u and dx as du/(6x)

this is the reason 1/6 appears in the 2nd exmple ...

for the same reasons

in the 3rd example

u = 5x^4 + 9

du = 20 x^3 dx

so now he has substituted 5x^4 + 9  as u and dx as du/(20x^3)

when ever the integral is of the form

U ' (x) * U(x) dx

we can write it as U(x) d (U(x))

in the 2nd example

it was x* sqrt ( 3x^2 - 5 ) dx

we can write it as 1/6 * 6x* sqrt ( 3x^2 - 5 ) dx

but 6x is what we get after differentiation of ( 3x^2 - 5 )

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