du=x^5+3x+C ?? what is that all abot
You add a constant to integral is no constraints on it. BUT NOT to the derivative
if
u = 5 x^4 + 3
then
du = 20 x^3 dx period, the end but that is not what you want to do anyway
Evaluate using substitution ∫(2x^5+6x)^3(5x^4+3)dx where b=0 and a=-1
I got to here and then got stuck
u=5x^4+3
du=x^5+3x+C
∫(2x^5+6x)^3 u 1/(x^5+3x+C)
∫(x^5+3x)^3 u
3 answers
lets' try
u = 2 x^5 + 6 x
then
du = (10 x^4 + 6)dx = 2 (5 x^4+3)dx
that looks more promising :)
u = 2 x^5 + 6 x
then
du = (10 x^4 + 6)dx = 2 (5 x^4+3)dx
that looks more promising :)
∫(2x^5+6x)^3(5x^4+3)dx
becomes
∫(u)^3 du/2
= (1/2) ∫u^3 du
BUT remember to change b = -1 to
2(-1)^5 +6(-6)
=-2 -36
= -38
becomes
∫(u)^3 du/2
= (1/2) ∫u^3 du
BUT remember to change b = -1 to
2(-1)^5 +6(-6)
=-2 -36
= -38
1 answer