Question

determine integral of 2x + 1/ x^2 + x dx

Answer 1

Let u=x^2+x
Then du=2x+1
So your integral becomes:
\[\int\limits \frac{du}{u}=\ln|u|+C=\ln|x^2+x|+C\]

Answer 2

Should read: du=2x+1 dx***

Answer 3

is that my final answer?

Answer 4

Yessir :)

Answer 5

how do you do integral 2y^3 e^2y4-2 dy

Answer 6

Okay:
\[\int\limits 2y^3 e^{2y^4-2}dy\]
Let u=2y^4-2
du=8y^3dy
so
(1/4)du=2y^3dy
So your integral becomes:
\[\frac{1}{4}\int\limits e^u du\]
What do you do from here?

Answer 7

ok I got 1/4e ^(2y^4-1) + C

Answer 8

Exactly :D

Answer 9

Any other?

Answer 10

yeah integral of xe^(2x squared) dx

Answer 11

du=4x dx

Answer 12

dx= 1/4 du

Answer 13

We don't have to complicate, we just have to simplify the fuction that the integer give to us:
\[\int\limits_{}^{}(((2x+1)\div(x^2))+x)dx\] \[2\int\limits_{}^{}(1/x) dx + \int\limits_{}^{}(1/x^2)dx + \int\limits_{}^{}xdx\] So, \[2\ln x-(1/x)+((x^2/2))+C\]
I hope i help you