Have the first four answers correct. just need with the last one. consider the integral ? 6x3(x4+1)dx. in the following, we will evaluate the integral using two methods. a. first, rewrite the integral by multiplying out the integrand: ? 6x3(x4+1)dx=? 6x^7+6x^3 dxthen evaluate the resulting integral term-by-term: ? 6x3(x4+1)dx= ((6x^8)/8)+((6x^4)/4)+cb. next, rewrite the integral using the substitution w=x4+1: ? 6x3(x4+1)dx=/4)*w) dwevaluate this integral (and back-substitute for w) to find the value of the original integral: ? 6x3(x4+1)dx= ((3(x^4+1)^2)/4)+cc. how are your expressions from parts (a) and (b) different? what is the difference between the two? (ignore the constant of from from a) =are both of the answers correct? (be sure you can explain why they are! )

Both of the solutions differ by their constant of integration.

Both of the answers are correct.

Step-by-step explanation:

Both of the answers are correct.

A. ∫6x³(x⁴+1)dx

if we apply the distributive property

∫6x³(x⁴+1)dx = ∫(6x⁷+ 6x³)dx = (6/8)x⁸ + (6/4)x⁴ + C = (3/4)x⁸ + (3/2)x⁴ + C

A. ∫6x³(x⁴+1)dx

If we use the substitution

w = x⁴+1

⇒ dw = 4x³dx  ⇒  x³dx = (1/4)dw

we have

∫6x³(x⁴+1)dx = 6∫w*(1/4)dw = (6/4)∫w dw = (3/2) (w²/2) + C = (3/4)w² + C

returning the change we get

∫6x³(x⁴+1)dx = (3/4)(x⁴+1)² + C

For these answer we can apply Notable Identities as follows

(x⁴+1)² = (x⁴)² + 2*x⁴*(1) + (1)² = x⁸ + 2x⁴ + 1

then

(3/4)(x⁴+1)² + C₂ = (3/4)(x⁸ + 2x⁴ + 1) + C = (3/4)*x⁸ + (3/2)*x⁴ + (3/4) + C

We can assume (3/4) + C as a constant, then

∫6x³(x⁴+1)dx = (3/4)*x⁸ + (3/2)*x⁴ + C

Both of the solutions differ by their constant of integration.

Both solutions are primitive of the same function, which must only fulfill a condition that is to differentiate into a constant, and in this case they fulfill it. It is not usual that using different methods of integration obtain different primitives, but it is not uncommon, there are quite a few known cases in which this happens and absolutely nothing happens. It is normal. Two functions that differ in a constant have the same derivative and therefore, both functions are primitive of that same derivative.

C is often forgotten in result.

honesty mood

step-by-step explanation: